Numerical on Refraction of Light for CBSE Class 10

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Numerical Problems on Refractive Index

Class 10 Science · Light – Reflection and Refraction
N

Refractive Index & Snell’s Law Practice Set 1

Apply the formulas $n = \frac{c}{v}$ and $\frac{\sin i}{\sin r} = \frac{n_2}{n_1}$
Ch 10 · Light
1
Light travels through water with a speed of $2.25 \times 10^8\text{ m/s}$. What is the refractive index of water?
Answer & Solution

Answer: $1.33$ (or $4/3$)

Explanation: The absolute refractive index ($n$) of a medium is the ratio of the speed of light in vacuum ($c$) to the speed of light in the medium ($v$). Note: $c = 3 \times 10^8\text{ m/s}$.

$$n = \frac{c}{v} = \frac{3 \times 10^8}{2.25 \times 10^8}$$
$$n = \frac{300}{225} = \frac{4}{3} \approx 1.33$$
2
Light travels from rarer medium 1 to a denser medium 2. The angle of incident and refraction are respectively $45^\circ$ and $30^\circ$. Calculate the (i) refractive index of second medium with respect to the first medium and (ii) refractive index of medium 1 with respect to the medium 2.
Answer & Solution

Answer: (i) $1.414$ (or $\sqrt{2}$) (ii) $0.707$ (or $1/\sqrt{2}$)

Explanation: Given $i = 45^\circ$ and $r = 30^\circ$.

(i) Refractive index of medium 2 w.r.t medium 1 ($n_{21}$) is given by Snell’s Law:

$$n_{21} = \frac{\sin i}{\sin r} = \frac{\sin 45^\circ}{\sin 30^\circ} = \frac{1/\sqrt{2}}{1/2}$$
$$n_{21} = \frac{2}{\sqrt{2}} = \sqrt{2} \approx 1.414$$

(ii) Refractive index of medium 1 w.r.t medium 2 ($n_{12}$) is the reciprocal of $n_{21}$:

$$n_{12} = \frac{1}{n_{21}} = \frac{1}{\sqrt{2}} \approx 0.707$$
3
A pond of depth 20 cm is filled with water of refractive index 4/3. Calculate apparent depth of the tank when viewed normally.
Answer & Solution

Answer: $15\text{ cm}$

Explanation: The formula relating real depth, apparent depth, and refractive index is:

$$n = \frac{\text{Real Depth}}{\text{Apparent Depth}}$$

Given $n = 4/3$ and Real Depth $= 20\text{ cm}$.

$$\frac{4}{3} = \frac{20}{\text{Apparent Depth}}$$
$$\text{Apparent Depth} = \frac{20 \times 3}{4} = 15\text{ cm}$$
4
How much time will light take to cross 2 mm thick glass pane if refractive index of glasses is 3/2?
Answer & Solution

Answer: $10^{-11}\text{ s}$

Explanation: First, calculate the speed of light in glass. $n_g = 1.5$ (or $3/2$).

$$v = \frac{c}{n_g} = \frac{3 \times 10^8}{3/2} = 2 \times 10^8\text{ m/s}$$

Next, convert the thickness of the glass to meters: $d = 2\text{ mm} = 2 \times 10^{-3}\text{ m}$.

$$t = \frac{\text{Distance}}{\text{Speed}} = \frac{2 \times 10^{-3}}{2 \times 10^8}$$
$$t = 1 \times 10^{-11}\text{ s}$$
5
Calculate speed of light in water of refractive index 4/3.
Answer & Solution

Answer: $2.25 \times 10^8\text{ m/s}$

Explanation: Using the formula $v = \frac{c}{n}$ where $c = 3 \times 10^8\text{ m/s}$.

$$v = \frac{3 \times 10^8}{4/3} = \frac{3 \times 10^8 \times 3}{4}$$
$$v = \frac{9 \times 10^8}{4} = 2.25 \times 10^8\text{ m/s}$$
6
A ray of light passes from air to glass (n = 1.5) at an angle of $30^\circ$. Calculate the angle of refraction.
Answer & Solution

Answer: $19.47^\circ$

Explanation: Refractive index of air is $n_1 \approx 1$. Glass is $n_2 = 1.5$. Angle of incidence $i = 30^\circ$. Applying Snell’s Law:

$$n_1 \sin i = n_2 \sin r \implies 1 \times \sin 30^\circ = 1.5 \times \sin r$$
$$0.5 = 1.5 \sin r \implies \sin r = \frac{0.5}{1.5} = \frac{1}{3} \approx 0.3333$$
$$r = \sin^{-1}(0.3333) \approx 19.47^\circ$$
7
A ray of light is incident on a glass slab at an angle of $45^\circ$. If refractive index of glass be 1.6, what is the angle of refraction?
Answer & Solution

Answer: $26.23^\circ$

Explanation: Given $i = 45^\circ$ and $n = 1.6$. Using Snell’s law ($n = \frac{\sin i}{\sin r}$):

$$1.6 = \frac{\sin 45^\circ}{\sin r} \implies \sin r = \frac{\sin 45^\circ}{1.6}$$
$$\sin r = \frac{0.7071}{1.6} \approx 0.4419$$
$$r = \sin^{-1}(0.4419) \approx 26.23^\circ$$
8
The refractive index of diamond is 2.47 and that of glass is 1.51. How much faster does light travel in glass than in diamond?
Answer & Solution

Answer: $1.636$ times faster

Explanation: Since $v = \frac{c}{n}$, the ratio of speeds is inversely proportional to the ratio of their refractive indices.

$$\text{Ratio} = \frac{v_{glass}}{v_{diamond}} = \frac{c / n_{glass}}{c / n_{diamond}} = \frac{n_{diamond}}{n_{glass}}$$
$$\text{Ratio} = \frac{2.47}{1.51} \approx 1.636$$

Light travels approximately 1.636 times faster in glass than it does in diamond.

N

Refractive Index & Real/Apparent Depth Practice Set 2

Apply relative refractive index and depth formulas
Ch 10 · Light
9
The refractive index of glycerine is 1.46. What is the speed of light in air if its speed in glycerine is $2.05 \times 10^8\text{ m/s}$?
Answer & Solution

Answer: $2.99 \times 10^8\text{ m/s}$ (approx $3 \times 10^8\text{ m/s}$)

Explanation: Given refractive index $n = 1.46$ and speed in glycerine $v = 2.05 \times 10^8\text{ m/s}$. The speed of light in air is practically equal to $c$ (speed of light in vacuum).

$$n = \frac{c}{v} \implies c = n \times v$$
$$c = 1.46 \times (2.05 \times 10^8) = 2.993 \times 10^8\text{ m/s}$$
10
The refractive index of glass is 1.6 and that of diamond is 2.4. Calculate (i) refractive index of diamond with respect to glass and (ii) refractive index of glass with respect to diamond.
Answer & Solution

Answer: (i) $1.5$ (ii) $0.67$

Explanation: Absolute RI of glass ($n_g$) = 1.6, diamond ($n_d$) = 2.4.

(i) RI of diamond w.r.t glass ($n_{dg}$):

$$n_{dg} = \frac{n_d}{n_g} = \frac{2.4}{1.6} = 1.5$$

(ii) RI of glass w.r.t diamond ($n_{gd}$):

$$n_{gd} = \frac{n_g}{n_d} = \frac{1.6}{2.4} = \frac{2}{3} \approx 0.67$$
11
A ray of light is travelling from glass to air. The angle of incidence in glass is $30^\circ$ and angle of refraction in air is $60^\circ$. What is the refractive index of glass w.r.t air?
Answer & Solution

Answer: $1.732$ (or $\sqrt{3}$)

Explanation: According to Snell’s law: $n_1 \sin i = n_2 \sin r$.
Here medium 1 is glass ($i = 30^\circ$) and medium 2 is air ($r = 60^\circ$).

$$n_g \sin 30^\circ = n_a \sin 60^\circ$$

The refractive index of glass w.r.t air ($n_{ga}$) is $\frac{n_g}{n_a}$:

$$n_{ga} = \frac{\sin 60^\circ}{\sin 30^\circ} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3} \approx 1.732$$
12
A ray of light is travelling from air to water. What is the angle of incidence in air, if angle of refraction in water is $45^\circ$? Take refractive index of water = 1.32.
Answer & Solution

Answer: $68.9^\circ$

Explanation: For light traveling from air ($n_1 = 1$) to water ($n_2 = 1.32$):

$$n_1 \sin i = n_2 \sin r$$
$$1 \times \sin i = 1.32 \times \sin 45^\circ$$
$$\sin i = 1.32 \times 0.7071 = 0.9333$$
$$i = \sin^{-1}(0.9333) \approx 68.96^\circ$$
13
A water tank appears to be 4 m deep when viewed from the top. If refractive index of water is 4/3, what is the actual depth of the tank?
Answer & Solution

Answer: $5.33\text{ m}$

Explanation: Given Apparent Depth = $4\text{ m}$ and Refractive Index $n = 4/3$.

$$n = \frac{\text{Real Depth}}{\text{Apparent Depth}}$$
$$\text{Real Depth} = n \times \text{Apparent Depth} = \frac{4}{3} \times 4 = \frac{16}{3}$$
$$\text{Real Depth} \approx 5.33\text{ m}$$
14
What is the real depth of a swimming pool when its bottom appears to be raised by 1 m? Given refractive index of water is 4/3.
Answer & Solution

Answer: $4\text{ m}$

Explanation: The shift or apparent rise ($d$) is given by the difference between Real Depth ($RD$) and Apparent Depth ($AD$).

$$d = RD – AD = RD – \frac{RD}{n} = RD \left(1 – \frac{1}{n}\right)$$

Given $d = 1\text{ m}$ and $n = 4/3$:

$$1 = RD \left(1 – \frac{1}{4/3}\right) = RD \left(1 – \frac{3}{4}\right)$$
$$1 = RD \left(\frac{1}{4}\right) \implies RD = 4\text{ m}$$
15
A jar 15 cm long is filled with a transparent liquid. When viewed from the top, its bottom appears to be 12 cm below. What is the refractive index of the liquid?
Answer & Solution

Answer: $1.25$

Explanation: The physical depth of the jar is the Real Depth $= 15\text{ cm}$. The depth it appears to be is the Apparent Depth $= 12\text{ cm}$.

$$n = \frac{\text{Real Depth}}{\text{Apparent Depth}} = \frac{15}{12}$$
$$n = \frac{5}{4} = 1.25$$

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