Ray Optics and Optical Instruments — NCERT Solutions

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

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).

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: $u < 0$

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 $2f$. ✓

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. ✓

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. ✓

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.