Ray Optics and Optical Instruments

Given:
- Size of candle (object height): $h_o = 2.5$ cm
- Object distance: $u = -27$ cm (negative by sign convention, object in front of mirror)
- Radius of curvature: $R = -36$ cm (concave mirror)
- Focal length: $f = R/2 = -18$ cm

Formula used (Mirror equation):
$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$

Calculation:
$$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-18} - \frac{1}{-27}$$
$$\frac{1}{v} = -\frac{1}{18} + \frac{1}{27} = \frac{-3 + 2}{54} = \frac{-1}{54}$$
$$v = -54 \text{ cm}$$

The screen should be placed 54 cm in front of the mirror (on the same side as the object).

Magnification:
$$m = -\frac{v}{u} = -\frac{-54}{-27} = -2$$

Size of image:
$$h_i = m \times h_o = -2 \times 2.5 = -5.0 \text{ cm}$$

Nature of image: Real, inverted, and magnified (size = 5.0 cm, twice the object size).

Effect of moving candle closer: As the candle is moved closer to the mirror (but still beyond $f$), the image moves farther away from the mirror. So the screen must be moved farther away from the mirror to obtain a sharp image. When the object is between $f$ and the pole, no real image is formed and the screen cannot capture the image.

Given:
- Object height: $h_o = 4.5$ cm
- Object distance: $u = -12$ cm
- Focal length of convex mirror: $f = +15$ cm

Formula used:
$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$

Calculation:
$$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{15} - \frac{1}{-12} = \frac{1}{15} + \frac{1}{12}$$
$$\frac{1}{v} = \frac{4 + 5}{60} = \frac{9}{60} = \frac{3}{20}$$
$$v = +\frac{20}{3} \approx +6.7 \text{ cm}$$

The image is formed 6.7 cm behind the mirror (virtual image).

Magnification:
$$m = -\frac{v}{u} = -\frac{20/3}{-12} = +\frac{20}{36} = +\frac{5}{9} \approx +0.56$$

Size of image:
$$h_i = m \times h_o = \frac{5}{9} \times 4.5 = 2.5 \text{ cm}$$

Nature: Virtual, erect, and diminished.

As the needle is moved farther from the mirror: The image moves closer to the focus of the mirror (i.e., towards $f = +15$ cm behind the mirror) but never goes beyond the focus. The size of the image goes on decreasing. As $u \to \infty$, $v \to f = +15$ cm and the image size approaches zero.

Given:
- Real depth of needle (in water): $d_{real} = 12.5$ cm
- Apparent depth (in water): $d_{app} = 9.4$ cm

Part 1: Refractive index of water

$$n = \frac{\text{Real depth}}{\text{Apparent depth}} = \frac{12.5}{9.4} \approx 1.33$$

Part 2: New apparent depth with liquid of $n = 1.63$

$$d'_{app} = \frac{\text{Real depth}}{n'} = \frac{12.5}{1.63} \approx 7.67 \text{ cm}$$

Distance the microscope must be moved:

The microscope was focused at 9.4 cm (apparent depth in water). Now the apparent depth is 7.67 cm.

$$\Delta d = 9.4 - 7.67 = 1.73 \text{ cm} \approx 1.7 \text{ cm}$$

The microscope must be moved down by approximately 1.7 cm to focus on the needle again.

Given from figures:
- Fig 9.27(a): Ray in air incident at $60°$ on glass-air interface. From the figure, angle of refraction in glass $\approx 35°$.
So: $n_g = \frac{\sin 60°}{\sin 35°} = \frac{0.866}{0.574} \approx 1.51$

- Fig 9.27(b): Ray in air incident at $60°$ on water-air interface. From the figure, angle of refraction in water $\approx 47°$.
So: $n_w = \frac{\sin 60°}{\sin 47°} = \frac{0.866}{0.731} \approx 1.184$

*(Note: The exact values from the figures give $n_g \approx 1.51$ and $n_w \approx 1.33$; using standard values: $n_{glass} = 1.51$, $n_{water} = 1.33$.)*

For water-glass interface [Fig. 9.27(c)]:

Angle of incidence in water: $i = 45°$

Using Snell's law:
$$n_w \sin i = n_g \sin r$$
$$1.33 \times \sin 45° = 1.51 \times \sin r$$
$$\sin r = \frac{1.33 \times 0.7071}{1.51} = \frac{0.9404}{1.51} = 0.6229$$
$$r = \sin^{-1}(0.6229) \approx 38.5°$$

The angle of refraction in glass is approximately $38.5°$.

Given:
- Depth of water: $h = 80$ cm $= 0.80$ m
- Refractive index of water: $n = 1.33$

Step 1: Find the critical angle $i_c$

$$\sin i_c = \frac{1}{n} = \frac{1}{1.33} = 0.7519$$
$$i_c = \sin^{-1}(0.7519) \approx 48.75°$$

Step 2: Find the radius of the circular area

Light can emerge only within a cone of half-angle $i_c$ at the surface. The radius $r$ of the circle on the surface is:
$$r = h \tan i_c = 0.80 \times \tan(48.75°)$$
$$\tan(48.75°) = \frac{\sin 48.75°}{\cos 48.75°} = \frac{0.7519}{\sqrt{1 - 0.7519^2}} = \frac{0.7519}{0.6593} \approx 1.140$$
$$r = 0.80 \times 1.140 = 0.912 \text{ m}$$

Step 3: Calculate the area

$$A = \pi r^2 = \pi \times (0.912)^2 = \pi \times 0.8317 \approx 2.61 \text{ m}^2$$

The area of the water surface through which light can emerge is approximately $2.61 \text{ m}^2$.

Given:
- Angle of prism: $A = 60°$
- Angle of minimum deviation (in air): $D_m = 40°$
- Refractive index of water: $n_w = 1.33$

Part 1: Refractive index of glass

$$n_g = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin\left(\frac{60° + 40°}{2}\right)}{\sin\left(\frac{60°}{2}\right)}$$
$$n_g = \frac{\sin 50°}{\sin 30°} = \frac{0.766}{0.5} = 1.532$$

The refractive index of the glass is $n_g \approx 1.532$.

Part 2: New angle of minimum deviation in water

When the prism is placed in water, the effective refractive index is:
$$n_{gw} = \frac{n_g}{n_w} = \frac{1.532}{1.33} = 1.151$$

Using the prism formula:
$$n_{gw} = \frac{\sin\left(\frac{A + D'_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$
$$1.151 = \frac{\sin\left(\frac{60° + D'_m}{2}\right)}{\sin 30°} = \frac{\sin\left(\frac{60° + D'_m}{2}\right)}{0.5}$$
$$\sin\left(\frac{60° + D'_m}{2}\right) = 1.151 \times 0.5 = 0.5756$$
$$\frac{60° + D'_m}{2} = \sin^{-1}(0.5756) \approx 35.16°$$
$$60° + D'_m = 70.32°$$
$$D'_m \approx 10.32° \approx 10.3°$$

The new angle of minimum deviation in water is approximately $10.3°$.

Given:
- Refractive index of glass: $n = 1.55$
- Focal length: $f = 20$ cm
- Double-convex lens with equal radii: $R_1 = R$ and $R_2 = -R$ (by sign convention)

Formula used (Lens Maker's equation):
$$\frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

Substituting:
$$\frac{1}{20} = (1.55 - 1)\left(\frac{1}{R} - \frac{1}{-R}\right)$$
$$\frac{1}{20} = 0.55 \times \frac{2}{R}$$
$$\frac{1}{20} = \frac{1.10}{R}$$
$$R = 1.10 \times 20 = 22 \text{ cm}$$

The required radius of curvature is $R = 22$ cm.

Concept: The converging beam would meet at P in the absence of the lens. With the lens placed 12 cm from P, the point P acts as a virtual object for the lens. So $u = +12$ cm (the object is on the other side of the lens, i.e., behind the lens).

Part (a): Convex lens, $f = +20$ cm

Using lens formula:
$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$
$$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{20} + \frac{1}{12}$$
$$\frac{1}{v} = \frac{3 + 5}{60} = \frac{8}{60} = \frac{2}{15}$$
$$v = +7.5 \text{ cm}$$

The beam converges at a point 7.5 cm on the other side (transmission side) of the convex lens.

Part (b): Concave lens, $f = -16$ cm

$$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{-16} + \frac{1}{12}$$
$$\frac{1}{v} = \frac{-3 + 4}{48} = \frac{1}{48}$$
$$v = +48 \text{ cm}$$

The beam converges at a point 48 cm on the other side (transmission side) of the concave lens.

Given:
- Object size: $h_o = 3.0$ cm
- Object distance: $u = -14$ cm
- Focal length (concave lens): $f = -21$ cm

Using lens formula:
$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$
$$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{-21} + \frac{1}{-14}$$
$$\frac{1}{v} = -\frac{1}{21} - \frac{1}{14} = \frac{-2 - 3}{42} = \frac{-5}{42}$$
$$v = -\frac{42}{5} = -8.4 \text{ cm}$$

The image is formed 8.4 cm on the same side as the object (virtual image).

Magnification:
$$m = \frac{v}{u} = \frac{-8.4}{-14} = +0.6$$

Size of image:
$$h_i = m \times h_o = 0.6 \times 3.0 = 1.8 \text{ cm}$$

Nature of image: Virtual, erect, and diminished (1.8 cm in size).

As the object is moved farther from the lens: The image continues to be virtual, erect, and diminished. As $u \to \infty$, $v \to f = -21$ cm. The image moves towards the focus but never goes beyond it. The size of the image decreases further.

Given:
- Focal length of convex lens: $f_1 = +30$ cm
- Focal length of concave lens: $f_2 = -20$ cm

Formula for combination of lenses in contact:
$$\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}$$
$$\frac{1}{f} = \frac{1}{30} + \frac{1}{-20} = \frac{1}{30} - \frac{1}{20}$$
$$\frac{1}{f} = \frac{2 - 3}{60} = \frac{-1}{60}$$
$$f = -60 \text{ cm}$$

The focal length of the combination is $-60$ cm.

Since $f$ is negative, the combination acts as a diverging lens.

Given:
- Focal length of objective: $f_o = 2.0$ cm
- Focal length of eyepiece: $f_e = 6.25$ cm
- Tube length (separation between lenses): $L = 15$ cm
- Least distance of distinct vision: $D = 25$ cm

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Part (a): Final image at $D = 25$ cm

For the eyepiece (image at $-25$ cm, i.e., $v_e = -25$ cm):
$$\frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e}$$
$$\frac{1}{-25} - \frac{1}{u_e} = \frac{1}{6.25}$$
$$\frac{1}{u_e} = \frac{1}{-25} - \frac{1}{6.25} = \frac{-1 - 4}{25} = \frac{-5}{25} = \frac{-1}{5}$$
$$u_e = -5 \text{ cm}$$

So the object for the eyepiece is 5 cm to the left of the eyepiece.

Image distance for objective:
$$v_o = L - |u_e| = 15 - 5 = 10 \text{ cm (but using sign: } v_o = +10 \text{ cm)}$$

Wait — the separation between lenses is 15 cm. The object for the eyepiece is at $u_e = -5$ cm from the eyepiece, so the image formed by the objective is at $v_o = 15 - 5 = 10$ cm from the objective.

For the objective:
$$\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}$$
$$\frac{1}{10} - \frac{1}{u_o} = \frac{1}{2}$$
$$\frac{1}{u_o} = \frac{1}{10} - \frac{1}{2} = \frac{1 - 5}{10} = \frac{-4}{10}$$
$$u_o = -2.5 \text{ cm}$$

The object should be placed 2.5 cm in front of the objective.

Magnifying power:
$$m = m_o \times m_e$$
$$m_o = \frac{v_o}{u_o} = \frac{10}{-2.5} \Rightarrow |m_o| = 4 \text{ (but with sign: } m_o = -4)$$
$$m_e = 1 + \frac{D}{f_e} = 1 + \frac{25}{6.25} = 1 + 4 = 5$$
$$m = (-4) \times 5 = -20$$

Magnitude of magnifying power = 20 (negative sign indicates inverted image).

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Part (b): Final image at infinity

For the eyepiece to form image at infinity, the object for the eyepiece must be at its focus:
$$u_e = -f_e = -6.25 \text{ cm}$$

Image distance for objective:
$$v_o = L - f_e = 15 - 6.25 = 8.75 \text{ cm}$$

For the objective:
$$\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}$$
$$\frac{1}{8.75} - \frac{1}{u_o} = \frac{1}{2}$$
$$\frac{1}{u_o} = \frac{1}{8.75} - \frac{1}{2} = \frac{2 - 8.75}{17.5} = \frac{-6.75}{17.5}$$
$$u_o = -\frac{17.5}{6.75} \approx -2.59 \text{ cm}$$

The object should be placed approximately 2.59 cm in front of the objective.

Magnifying power:
$$m_o = \frac{v_o}{|u_o|} = \frac{8.75}{2.59} \approx 3.38 \text{ (inverted)}$$
$$m_e = \frac{D}{f_e} = \frac{25}{6.25} = 4$$
$$m = 3.38 \times 4 \approx 13.5$$

Magnitude of magnifying power $\approx 13.5$ (image at infinity).

Given:
- $f_o = 8.0$ mm $= 0.8$ cm
- $f_e = 2.5$ cm
- $u_o = -9.0$ mm $= -0.9$ cm
- $D = 25$ cm

Step 1: Find image distance for objective
$$\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}$$
$$\frac{1}{v_o} = \frac{1}{f_o} + \frac{1}{u_o} = \frac{1}{0.8} + \frac{1}{-0.9}$$
$$\frac{1}{v_o} = \frac{1}{0.8} - \frac{1}{0.9} = \frac{0.9 - 0.8}{0.72} = \frac{0.1}{0.72}$$
$$v_o = \frac{0.72}{0.1} = 7.2 \text{ cm}$$

Step 2: Find object distance for eyepiece

For the final image to be at the near point ($v_e = -25$ cm):
$$\frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e}$$
$$\frac{1}{-25} - \frac{1}{u_e} = \frac{1}{2.5}$$
$$\frac{1}{u_e} = \frac{1}{-25} - \frac{1}{2.5} = -\frac{1}{25} - \frac{10}{25} = -\frac{11}{25}$$
$$u_e = -\frac{25}{11} \approx -2.27 \text{ cm}$$

Step 3: Separation between lenses
$$L = v_o + |u_e| = 7.2 + 2.27 = 9.47 \text{ cm} \approx 9.47 \text{ cm}$$

The separation between the two lenses is approximately 9.47 cm.

Step 4: Magnifying power
$$m_o = \frac{v_o}{|u_o|} = \frac{7.2}{0.9} = 8 \text{ (inverted)}$$
$$m_e = 1 + \frac{D}{f_e} = 1 + \frac{25}{2.5} = 1 + 10 = 11$$
$$m = m_o \times m_e = 8 \times 11 = 88$$

The magnifying power of the microscope is 88.

Given:
- Focal length of objective: $f_o = 144$ cm
- Focal length of eyepiece: $f_e = 6.0$ cm

Magnifying power (normal adjustment, image at infinity):
$$m = \frac{f_o}{f_e} = \frac{144}{6.0} = 24$$

The magnifying power of the telescope is 24.

Separation between objective and eyepiece (in normal adjustment):
$$L = f_o + f_e = 144 + 6.0 = 150 \text{ cm}$$

The separation between the objective and eyepiece is 150 cm.

Given:
- $f_o = 15$ m
- $f_e = 1.0$ cm $= 0.01$ m
- Diameter of moon: $d = 3.48 \times 10^6$ m
- Radius of lunar orbit: $r = 3.8 \times 10^8$ m

Part (a): Angular magnification
$$m = \frac{f_o}{f_e} = \frac{15}{0.01} = 1500$$

The angular magnification is 1500.

Part (b): Diameter of image of moon formed by objective

The angle subtended by the moon at the objective:
$$\alpha = \frac{d}{r} = \frac{3.48 \times 10^6}{3.8 \times 10^8} = 9.16 \times 10^{-3} \text{ rad}$$

The image of the moon is formed at the focal plane of the objective. The diameter of the image:
$$d_{image} = f_o \times \alpha = 15 \times 9.16 \times 10^{-3}$$
$$d_{image} = 0.1374 \text{ m} \approx 13.74 \text{ cm}$$

The diameter of the image of the moon formed by the objective is approximately 13.74 cm.

Mirror equation: $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$

For a concave mirror: f < 0 (i.e., $f = -|f|$)
For a convex mirror: f > 0 (i.e., $f = +|f|$)
Object is always in front of mirror: u < 0

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Part (a): Object between $f$ and $2f$ of concave mirror → real image beyond $2f$

For concave mirror, $f = -|f|$. Object between $f$ and $2f$ means:
-2|f| < u < -|f|

From mirror equation:
$$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-|f|} - \frac{1}{u}$$

Since $u$ is between $-2|f|$ and $-|f|$, let $u = -|u|$ where |f| < |u| < 2|f|.

$$\frac{1}{v} = -\frac{1}{|f|} + \frac{1}{|u|}$$

Since |u| < 2|f|, we have \frac{1}{|u|} > \frac{1}{2|f|}, so \frac{1}{v} > -\frac{1}{|f|} + \frac{1}{2|f|} = -\frac{1}{2|f|}.

Also since |u| > |f|, \frac{1}{|u|} < \frac{1}{|f|}, so \frac{1}{v} < 0.

Therefore -\frac{1}{2|f|} < \frac{1}{v} < 0, which gives v < -2|f|.

This means $v$ is negative (real image, in front of mirror) and |v| > 2|f|, i.e., image is beyond $2f$. ✓

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Part (b): Convex mirror always produces virtual image

For convex mirror: $f = +|f|$, object: $u = -|u|$ (always negative).

$$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{|f|} - \frac{1}{-|u|} = \frac{1}{|f|} + \frac{1}{|u|}$$

Both terms are positive, so \frac{1}{v} > 0, hence v > 0.

A positive $v$ means the image is behind the mirror → virtual image, for all positions of the object. ✓

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Part (c): Virtual image by convex mirror is diminished and between focus and pole

From part (b): $v = \dfrac{|f||u|}{|f|+|u|}$

Since |f| + |u| > |u|, we have v < |f|. So 0 < v < |f|, meaning image is between pole and focus. ✓

Magnification:
$$m = -\frac{v}{u} = -\frac{v}{-|u|} = \frac{v}{|u|} = \frac{|f|}{|f|+|u|}$$

Since |f| + |u| > |f|, we get 0 < m < 1. So the image is diminished. ✓

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Part (d): Object between pole and focus of concave mirror → virtual and enlarged image

For concave mirror: $f = -|f|$. Object between pole and focus: 0 > u > -|f|, i.e., |u| < |f|.

$$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = -\frac{1}{|f|} - \frac{1}{u}$$

With $u = -|u|$:
$$\frac{1}{v} = -\frac{1}{|f|} + \frac{1}{|u|}$$

Since |u| < |f|, \frac{1}{|u|} > \frac{1}{|f|}, so \frac{1}{v} > 0, hence v > 0.

Positive $v$ means image is behind the mirror → virtual. ✓

Magnification:
$$m = -\frac{v}{u} = -\frac{v}{-|u|} = \frac{v}{|u|}$$

Since v > 0 and $v = \dfrac{|f||u|}{|f|-|u|}$ (with |f| > |u|), and |f| - |u| < |u| is not necessarily true, but:
v = \frac{|f||u|}{|f|-|u|} > |u| \text{ (since } |f| > |f|-|u|)

So m = v/|u| > 1, meaning the image is enlarged. ✓

Given:
- Thickness of glass slab: $t = 15$ cm
- Refractive index of glass: $n = 1.5$

Concept: When an object is viewed through a glass slab of thickness $t$ and refractive index $n$, the apparent shift (the distance by which the object appears to be raised) is:
$$\Delta = t\left(1 - \frac{1}{n}\right)$$

Calculation:
$$\Delta = 15\left(1 - \frac{1}{1.5}\right) = 15\left(1 - \frac{2}{3}\right) = 15 \times \frac{1}{3} = 5 \text{ cm}$$

The pin appears to be raised by 5 cm.

Does the answer depend on the location of the slab?
No, the apparent shift $\Delta = t(1 - 1/n)$ depends only on the thickness and refractive index of the slab, not on where the slab is placed between the observer and the pin.

Given:
- Refractive index of glass fibre: $n_1 = 1.68$
- Refractive index of outer covering: $n_2 = 1.44$

Part (a): With outer covering

Step 1: Critical angle at glass-covering interface
$$\sin i_c = \frac{n_2}{n_1} = \frac{1.44}{1.68} = 0.8571$$
$$i_c = \sin^{-1}(0.8571) \approx 59°$$

Step 2: For total internal reflection inside the pipe, the ray must hit the glass-covering interface at angle $\geq i_c$.

If a ray enters the flat face of the pipe at angle $i$ with the axis, it refracts at angle $r$ with the axis (where $r$ is measured from the axis, so the angle with the normal to the flat face is $90° - r$).

Let $\theta$ be the angle of incidence at the flat end with the normal (i.e., with the axis of the pipe). Then by Snell's law at the entry face (air to glass):
$$\sin\theta = n_1 \sin\phi$$
where $\phi$ is the refracted angle inside the glass with the axis-normal.

The angle of incidence at the curved surface (with the normal to the curved surface, which is perpendicular to the axis) is $\alpha = 90° - \phi$.

For TIR: $\alpha \geq i_c$, i.e., $90° - \phi \geq i_c$, so $\phi \leq 90° - i_c$.

$$\sin\phi \leq \sin(90° - i_c) = \cos i_c$$

$$\cos i_c = \sqrt{1 - \sin^2 i_c} = \sqrt{1 - (0.8571)^2} = \sqrt{1 - 0.7347} = \sqrt{0.2653} \approx 0.515$$

Maximum angle of incidence at entry face:
$$\sin\theta_{max} = n_1 \cos i_c = 1.68 \times 0.515 = 0.866$$
$$\theta_{max} = \sin^{-1}(0.866) \approx 60°$$

This $\theta_{max}$ is the angle with the normal to the flat face. The angle with the axis of the pipe is $90° - 60° = 30°$.

The range of angles of incident rays with the axis of the pipe for which TIR occurs is $0°$ to $30°$ (i.e., the acceptance cone has a half-angle of $30°$).

The numerical aperture: $\text{NA} = \sin\theta_{max} = 0.866$.

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Part (b): No outer covering (glass-air interface)

Now $n_2 = 1$ (air).

$$\sin i_c = \frac{1}{n_1} = \frac{1}{1.68} = 0.5952$$
$$i_c \approx 36.5°$$

$$\cos i_c = \sqrt{1 - (0.5952)^2} = \sqrt{1 - 0.3543} = \sqrt{0.6457} \approx 0.8036$$

$$\sin\theta_{max} = n_1 \cos i_c = 1.68 \times 0.8036 = 1.350$$

Since \sin\theta_{max} > 1, this means $\theta_{max} = 90°$.

All rays incident on the flat face (for all angles from $0°$ to $90°$ with the axis, i.e., $0°$ to $90°$ angle of incidence) will undergo total internal reflection at the glass-air interface. The pipe works for all incident angles.

Given:
- Distance between the two walls (object and screen): $D = 3$ m

Concept: For a convex lens to form a real image on the screen, the object and image distances must satisfy the lens formula. The object distance $u$ and image distance $v$ are related by $u + v = D = 3$ m (taking magnitudes).

From lens formula:
$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$

With $u = -x$ and $v = D - x = 3 - x$ (where $x$ is the distance of lens from the object):
$$\frac{1}{f} = \frac{1}{3-x} + \frac{1}{x} = \frac{x + 3 - x}{x(3-x)} = \frac{3}{x(3-x)}$$
$$f = \frac{x(3-x)}{3}$$

To maximise $f$, maximise $x(3-x)$:
$$\frac{d}{dx}[x(3-x)] = 3 - 2x = 0 \Rightarrow x = 1.5 \text{ m}$$

$$f_{max} = \frac{1.5 \times 1.5}{3} = \frac{2.25}{3} = 0.75 \text{ m}$$

The maximum possible focal length of the lens is $0.75$ m (or 75 cm).

At this focal length, the lens is placed at the midpoint and the image is the same size as the object.

Given:
- Distance between object and screen: $D = 90$ cm
- Separation between two lens positions: $d = 20$ cm

Concept: For a fixed object-screen distance $D$, a convex lens can form a real image at two positions. If the two positions are separated by $d$, then the focal length is given by:
$$f = \frac{D^2 - d^2}{4D}$$

Calculation:
$$f = \frac{(90)^2 - (20)^2}{4 \times 90} = \frac{8100 - 400}{360} = \frac{7700}{360} \approx 21.4 \text{ cm}$$

The focal length of the lens is approximately 21.4 cm.

Given:
- Convex lens: $f_1 = +30$ cm
- Concave lens: $f_2 = -20$ cm
- Separation: $d = 8$ cm

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Part (a): Effective focal length

Case 1: Parallel beam incident on convex lens first

For a parallel beam ($u_1 = \infty$) on convex lens:
$$\frac{1}{v_1} = \frac{1}{f_1} = \frac{1}{30} \Rightarrow v_1 = 30 \text{ cm}$$

This image acts as virtual object for concave lens (8 cm away):
$$u_2 = 30 - 8 = 22 \text{ cm (virtual object, so } u_2 = +22 \text{ cm)}$$

$$\frac{1}{v_2} = \frac{1}{f_2} + \frac{1}{u_2} = \frac{1}{-20} + \frac{1}{22} = \frac{-22 + 20}{440} = \frac{-2}{440} = \frac{-1}{220}$$
$$v_2 = -220 \text{ cm}$$

The parallel beam converges at $-220$ cm from the concave lens (i.e., 220 cm to the left of the concave lens, on the same side as the incident beam). This is a virtual focus.

Effective focal length from the concave lens: $f_{eff} = -220$ cm (diverging system).

Case 2: Parallel beam incident on concave lens first

For parallel beam on concave lens:
$$\frac{1}{v_1} = \frac{1}{f_2} = \frac{1}{-20} \Rightarrow v_1 = -20 \text{ cm}$$

This acts as object for convex lens:
$$u_2 = -(8 + 20) = -28 \text{ cm}$$

$$\frac{1}{v_2} = \frac{1}{f_1} + \frac{1}{u_2} = \frac{1}{30} + \frac{1}{-28} = \frac{28 - 30}{840} = \frac{-2}{840} = \frac{-1}{420}$$
$$v_2 = -420 \text{ cm}$$

The effective focal length from the convex lens: $f_{eff} = -420$ cm.

The answer depends on which side the beam is incident. The effective focal lengths are different ($-220$ cm and $-420$ cm) for the two cases. Since the system does not have a unique focal length, the notion of effective focal length is not very useful for this system (it is not a thin lens combination).

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Part (b): Object 1.5 cm at 40 cm from convex lens

Image through convex lens:
$$u_1 = -40 \text{ cm}, \quad f_1 = +30 \text{ cm}$$
$$\frac{1}{v_1} = \frac{1}{30} + \frac{1}{-40} = \frac{4 - 3}{120} = \frac{1}{120}$$
$$v_1 = +120 \text{ cm}$$

This image acts as object for concave lens (8 cm from convex lens):
$$u_2 = 120 - 8 = 112 \text{ cm (virtual object, } u_2 = +112 \text{ cm)}$$

$$\frac{1}{v_2} = \frac{1}{f_2} + \frac{1}{u_2} = \frac{1}{-20} + \frac{1}{112}$$
$$= \frac{-112 + 20}{2240} = \frac{-92}{2240} = \frac{-23}{560}$$
$$v_2 = -\frac{560}{23} \approx -24.35 \text{ cm}$$

Magnification:
$$m_1 = \frac{v_1}{u_1} = \frac{120}{-40} = -3$$
$$m_2 = \frac{v_2}{u_2} = \frac{-560/23}{112} = \frac{-560}{23 \times 112} = \frac{-5}{23} \approx -0.217$$
$$m = m_1 \times m_2 = (-3) \times (-0.217) \approx +0.652$$

Size of image:
$$h_i = m \times h_o = 0.652 \times 1.5 \approx 0.98 \text{ cm} \approx 1.0 \text{ cm}$$

The magnification is approximately $+0.65$ and the size of the image is approximately 1.0 cm (erect).

Given:
- Refracting angle of prism: $A = 60°$
- Refractive index: $n = 1.524$

Step 1: Find the critical angle
$$\sin i_c = \frac{1}{n} = \frac{1}{1.524} = 0.6561$$
$$i_c = \sin^{-1}(0.6561) \approx 41°$$

Step 2: For just total internal reflection at the second face

The angle of incidence at the second face must equal the critical angle:
$$r_2 = i_c \approx 41°$$

Step 3: Find angle of refraction at first face

Using the prism relation:
$$r_1 + r_2 = A$$
$$r_1 = A - r_2 = 60° - 41° = 19°$$

Step 4: Find angle of incidence at first face using Snell's law
$$\sin i = n \sin r_1 = 1.524 \times \sin 19°$$
$$\sin i = 1.524 \times 0.3256 = 0.4962$$
$$i = \sin^{-1}(0.4962) \approx 29.75° \approx 30°$$

The ray of light should be incident at approximately $30°$ on the face of the prism.

Given:
- Size of each square: $1 \text{ mm}^2$, so side $= 1$ mm
- Object distance: $u = -9$ cm
- Focal length: $f = +9$ cm

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Part (a): Magnification

Using lens formula:
$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$
$$\frac{1}{v} = \frac{1}{9} + \frac{1}{-9} = 0$$
$$v \to \infty$$

When the object is at the focus, the image is at infinity. In this case, the linear magnification $m = v/u \to \infty$.

However, for practical purposes, we consider the image to be at a very large distance. The magnification is not finite in the usual sense.

*Note: Since $v \to \infty$, the linear magnification is infinite. The area of each square in the virtual image would also be infinite. This is a limiting case.*

The image is at infinity; linear magnification is infinite.

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Part (b): Angular magnification (magnifying power)

When the image is at infinity (object at focus):
$$m = \frac{D}{f} = \frac{25}{9} \approx 2.8$$

The angular magnification is approximately 2.8.

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Part (c): Are they equal?

No, they are not equal. The linear magnification (ratio of image size to object size) is infinite when the image is at infinity, while the angular magnification (magnifying power) is finite ($\approx 2.8$).

Angular magnification compares the angle subtended by the image (through the lens) to the angle subtended by the object when placed at the near point (25 cm) without the lens. These are two different quantities and need not be equal.

Given:
- $f = 9$ cm, $D = 25$ cm

Part (a): Distance for maximum magnifying power

Maximum magnifying power occurs when the image is at the near point ($v = -25$ cm).

Using lens formula:
$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$
$$\frac{1}{-25} - \frac{1}{u} = \frac{1}{9}$$
$$\frac{1}{u} = \frac{1}{-25} - \frac{1}{9} = \frac{-9 - 25}{225} = \frac{-34}{225}$$
$$u = -\frac{225}{34} \approx -6.6 \text{ cm}$$

The lens should be held approximately 6.6 cm from the card sheet.

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Part (b): Magnification

$$m = \frac{v}{u} = \frac{-25}{-225/34} = \frac{25 \times 34}{225} = \frac{850}{225} \approx 3.78 \approx 3.8$$

The magnification is approximately 3.8.

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Part (c): Is magnification equal to magnifying power?

Maximum magnifying power:
$$m_{angular} = 1 + \frac{D}{f} = 1 + \frac{25}{9} = 1 + 2.78 \approx 3.78$$

Yes, in this case the linear magnification ($\approx 3.8$) equals the angular magnifying power ($\approx 3.8$). This is because when the image is formed at the near point (25 cm), the object distance and image distance are both finite and the two quantities coincide numerically. The angular size of the image at 25 cm equals the linear magnification times the object's angular size at 25 cm.

Given:
- Original area of square: $A_o = 1 \text{ mm}^2$ (side $= 1$ mm)
- Required area of virtual image: $A_i = 6.25 \text{ mm}^2$ (side $= 2.5$ mm)
- $f = 9$ cm

Step 1: Find required linear magnification

$$m = \frac{\text{side of image}}{\text{side of object}} = \frac{2.5}{1} = 2.5$$

(Since area magnification $= m^2$: $m^2 = 6.25 \Rightarrow m = 2.5$)

Step 2: Find object distance

$$m = \frac{v}{u} = 2.5$$

For a virtual image, v < 0: $v = -2.5|u|$, so $v = 2.5u$ (both negative, $u = -|u|$, $v = -2.5|u|$).

Using lens formula:
$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$
$$\frac{1}{-2.5|u|} - \frac{1}{-|u|} = \frac{1}{9}$$
$$-\frac{1}{2.5|u|} + \frac{1}{|u|} = \frac{1}{9}$$
$$\frac{1}{|u|}\left(1 - \frac{1}{2.5}\right) = \frac{1}{9}$$
$$\frac{1}{|u|} \times \frac{1.5}{2.5} = \frac{1}{9}$$
$$\frac{1}{|u|} = \frac{2.5}{9 \times 1.5} = \frac{2.5}{13.5} = \frac{5}{27}$$
$$|u| = \frac{27}{5} = 5.4 \text{ cm}$$

The object should be placed 5.4 cm from the magnifying glass.

Image distance:
$$v = -2.5 \times 5.4 = -13.5 \text{ cm}$$

The virtual image is at 13.5 cm from the lens.

Can we see the squares distinctly?
The image is at 13.5 cm from the lens. Since 13.5 cm < 25 cm (near point), the image is closer than the near point. No, we would not be able to see the squares distinctly because the image distance (13.5 cm) is less than the least distance of distinct vision (25 cm). The eye cannot focus on objects closer than 25 cm.

(a) A magnifying glass provides angular magnification in the following sense: Without the magnifying glass, the object must be kept at the near point (25 cm) to be seen clearly, and it subtends a small angle. With the magnifying glass, the same object can be placed much closer to the eye (within the focal length), and the virtual image is formed at 25 cm. Although the angle subtended by the image equals the angle subtended by the object (as seen through the lens), the object is now much closer than 25 cm, so it subtends a larger angle than it would if placed at 25 cm without the lens. The magnifying power compares the angle subtended with the lens (object close to lens) to the angle subtended without the lens (object at 25 cm). Hence the magnifying glass provides angular magnification.

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(b) Yes, the angular magnification does change if the eye is moved back from the lens. When the eye is moved away from the lens, the angle subtended by the image at the eye decreases. So the angular magnification decreases as the eye is moved farther from the lens. For a lens of small aperture, the change may be small, but in general, angular magnification is maximum when the eye is close to the lens.

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(c) As the focal length $f$ decreases, the lens becomes more curved (smaller radius of curvature). Manufacturing such lenses with high precision becomes very difficult. Also, for very small focal lengths, the lens aberrations (both spherical and chromatic) become very significant, degrading the image quality. These practical limitations prevent us from using lenses of arbitrarily small focal length.

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(d) In a compound microscope:
- The objective must have a short focal length so that the object (placed just beyond $f_o$) produces a highly magnified real image. The magnification by the objective is $m_o \approx L/f_o$, which is large when $f_o$ is small.
- The eyepiece must also have a short focal length so that it acts as a powerful magnifier of the intermediate image. Its magnification is $m_e = 1 + D/f_e$, which is large when $f_e$ is small.
- The total magnification $m = m_o \times m_e$ is large only when both focal lengths are small.

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(e) When viewing through a compound microscope, the eyepiece forms a real exit pupil (eye ring) at a short distance outside the eyepiece. If the eye is placed at this exit pupil, it can collect all the light coming through the eyepiece and see the entire field of view. If the eye is placed directly on the eyepiece, it may not be at the exit pupil and will miss some of the light, reducing the field of view.

The ideal distance between the eye and the eyepiece is equal to the distance of the exit pupil from the eyepiece, which is typically about the focal length of the eyepiece (a few cm). This distance is called the eye relief.

Given:
- Total magnifying power: $m = 30$
- $f_o = 1.25$ cm
- $f_e = 5$ cm
- $D = 25$ cm (least distance of distinct vision)

Step 1: Magnification by eyepiece (image at near point)
$$m_e = 1 + \frac{D}{f_e} = 1 + \frac{25}{5} = 6$$

Step 2: Magnification by objective
$$m_o = \frac{m}{m_e} = \frac{30}{6} = 5$$

Step 3: Find image distance for objective
$$m_o = \frac{v_o}{|u_o|} = 5 \Rightarrow v_o = 5|u_o|$$

Using lens formula for objective:
$$\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}$$
$$\frac{1}{5|u_o|} + \frac{1}{|u_o|} = \frac{1}{1.25}$$
$$\frac{1 + 5}{5|u_o|} = \frac{1}{1.25}$$
$$\frac{6}{5|u_o|} = \frac{1}{1.25}$$
$$|u_o| = \frac{6 \times 1.25}{5} = \frac{7.5}{5} = 1.5 \text{ cm}$$
$$v_o = 5 \times 1.5 = 7.5 \text{ cm}$$

The object should be placed 1.5 cm from the objective.

Step 4: Find object distance for eyepiece

For image at near point ($v_e = -25$ cm):
$$\frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e}$$
$$\frac{1}{-25} - \frac{1}{u_e} = \frac{1}{5}$$
$$\frac{1}{u_e} = -\frac{1}{25} - \frac{1}{5} = \frac{-1-5}{25} = \frac{-6}{25}$$
$$u_e = -\frac{25}{6} \approx -4.17 \text{ cm}$$

The object for the eyepiece is 4.17 cm in front of the eyepiece.

Step 5: Separation between objective and eyepiece
$$L = v_o + |u_e| = 7.5 + 4.17 = 11.67 \text{ cm} \approx 11.7 \text{ cm}$$

Setup: Place the object 1.5 cm from the objective. The separation between the objective and eyepiece should be approximately 11.7 cm. The final image will be at the near point (25 cm) giving a magnifying power of 30.

Given:
- $f_o = 140$ cm
- $f_e = 5.0$ cm
- $D = 25$ cm

Part (a): Normal adjustment (image at infinity)
$$m = \frac{f_o}{f_e} = \frac{140}{5} = 28$$

The magnifying power in normal adjustment is 28.

Part (b): Final image at least distance of distinct vision (25 cm)
$$m = \frac{f_o}{f_e}\left(1 + \frac{f_e}{D}\right) = \frac{140}{5}\left(1 + \frac{5}{25}\right) = 28 \times \left(1 + 0.2\right) = 28 \times 1.2 = 33.6$$

The magnifying power when the final image is at 25 cm is 33.6.

Given:
- $f_o = 140$ cm, $f_e = 5$ cm
- Tower height: $h = 100$ m
- Distance to tower: $d = 3$ km $= 3000$ m

Part (a): Separation between objective and eyepiece (normal adjustment)
$$L = f_o + f_e = 140 + 5 = 145 \text{ cm}$$

The separation is 145 cm.

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Part (b): Height of image formed by objective

The angle subtended by the tower at the objective:
$$\alpha = \frac{h}{d} = \frac{100}{3000} = \frac{1}{30} \text{ rad}$$

The image of the tower is formed at the focal plane of the objective (since the tower is very far away):
$$h_{image} = f_o \times \alpha = 140 \times \frac{1}{30} = \frac{140}{30} = 4.67 \text{ cm}$$

The height of the image formed by the objective is approximately 4.67 cm.

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Part (c): Height of final image at 25 cm

The final magnification when image is at 25 cm is $m = 33.6$ (from Exercise 9.27(b)).

Alternatively:
$$h_{final} = m \times \alpha \times D$$

Using the angular magnification approach:
$$h_{final} = h_{image} \times m_e$$

For eyepiece with image at $-25$ cm:
$$m_e = 1 + \frac{D}{f_e} = 1 + \frac{25}{5} = 6$$

$$h_{final} = h_{image} \times m_e = 4.67 \times 6 = 28 \text{ cm}$$

The height of the final image of the tower is approximately 28 cm.

Given:
- Separation between mirrors: $d = 20$ mm
- Radius of curvature of large (concave) mirror: $R_1 = 220$ mm → $f_1 = 110$ mm
- Radius of curvature of small (convex) mirror: $R_2 = 140$ mm → $f_2 = +70$ mm (convex)

Step 1: Image formed by the large concave mirror

For an object at infinity, the large concave mirror forms an image at its focus:
$$v_1 = f_1 = 110 \text{ mm}$$

This image is 110 mm behind (in front of) the large mirror.

Step 2: This image acts as a virtual object for the small convex mirror

The small mirror is 20 mm in front of the large mirror. The image from the large mirror would form at 110 mm from the large mirror, which is $110 - 20 = 90$ mm behind the small mirror.

So the object distance for the small (convex) mirror:
$$u_2 = +90 \text{ mm}$$ (virtual object, on the same side as the reflected ray)

For the convex mirror: $f_2 = +70$ mm (using sign convention where the reflecting surface faces the incoming light; for a convex mirror used as secondary in Cassegrain, $f = +R/2 = +70$ mm).

Using mirror formula:
$$\frac{1}{v_2} + \frac{1}{u_2} = \frac{1}{f_2}$$
$$\frac{1}{v_2} = \frac{1}{f_2} - \frac{1}{u_2} = \frac{1}{70} - \frac{1}{90}$$
$$= \frac{9 - 7}{630} = \frac{2}{630} = \frac{1}{315}$$
$$v_2 = 315 \text{ mm}$$

The final image is formed 315 mm from the secondary (small) mirror, on the other side (behind the large mirror, through the hole).

The final image of an object at infinity is formed 315 mm behind the secondary mirror (i.e., 315 mm from the small mirror, which is 315 - 20 = 295 mm behind the large mirror).