## Optics

## Refraction Of Light Short Questions And Answers

**Question 1. A prism is made of glass of unknown refractive index A parallel beam of light incident on face of the prism. By rotating the prism, the angle of minimum deviation la j measured to be 40° What is the refractive Index of the material of the prism? If the prism is placed In water j (refractive index 1.33 ). predict the new angle of minimum j deviation of a parallel! beam of light Refracting angle of the prism is 60°**

**Answer: **

Refrractive index of the material of the prism

μ = \(\frac{\sin \frac{A+8}{2}}{\sin \frac{A}{2}}=\frac{\sin \frac{60^{\circ}+40^{\circ}}{2}}{\sin \frac{60^{\circ}}{2}}\)

= \(\frac{\sin 50^{\circ}}{\sin 30^{\circ}}\)

= 1.53

When placed in water,

_{w} μ_{g}= _{a}μ_{g} /_{a}μ_{w}

= \(\frac{1.53}{1.33}\)

_{w} μ_{g}= \({ }\frac{\sin \frac{A+\delta_m^{\prime}}{2}}{\sin \frac{A}{2}}\)

Or, 1.15\(\sin \frac{60^{\circ}}{2}=\sin \frac{60^{\circ}+\delta_m^{\prime}}{2}\)

∴ δ’_{m}= 10.2°

= 10. 12′

**Question 2. A small pin fixed on the table and top of it is viewed from above from a distance of 50 cm. By what distance would the tire pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab held parallel to the table?**

The refractive index of glass = 1.5 . Does the answer depend on the location of the slab? The apparent displacement of the pin

x = d\(\left(1-\frac{1}{\mu}\right)\)

[ d= apparent depth and = refractive index]

= 15 (1-\(\left(1-\frac{1}{1.5}\right)\) = 5 cm

Fors mall angles of Incidence the answer will not depend on the position of the glass

**Question 3. A cross glass fibre with a refractive Index 1.68. The outer covering of the pipe is made of material of a refractive index of 1.44. What Is the tango of the angles of Incident rays with the axis of the pipe for which the total internal reflection inside the pipe takes place as shown In the figure? ****What is the answer If there is no outer covering of the pipe? If them is no covering, then**

If there is no covering, then

⇒ \(i^{\prime}=\sin ^{-1}\left(\frac{1}{1.68}\right) \approx 36.5^{\circ}\)

We know, \(\frac{\sin t}{\sin r}=\mu\)

∴ sin i = sin 53.5° × 1.65 a 80.3 × 1.65 = 1.33 which is absurd as sin i_{max } = sin 90° = 1

A mass must be less than 53.5° which means

0°<K90° Thus total internal reflection will take plate for any angle of incidence ranging from to 90°

**Question 4. A diver underwater looks obliquely at a fisherman standing on the bank of a lake. Would the fisherman look taller or shorter than what he actually Is?
Answer:**

The displacement of the Image of the head of the fisherman The displacement of the Image of the head of the fisherman

**Question 5. At what angle should a ray of light be Incident on the face of a prism of refractive angle 60° so that it Just suffers total internal reflection at the other face? The refractive index of the material of the prism Is 1.524. **

**Answer:**

The limiting value of the angle of incidence,

i = \(\sin ^{-1}\left[\sin A \sqrt{\mu^2-1}-\cos A\right]\)

= \(\sin ^{-1}\left[\sin 60^{\circ}\left(\sqrt{1.524^2-1}\right)-\cos 60^{\circ}\right]\)

= 29.73 30

**Question 6. Calculate the speed of light in a medium whose critical angle is 45s. Mention two practical applications of optical fibre. **

If is the critical angle of any medium with respect to vacuum or air and // is the refractive index ofthe medium

⇒ \(\sin \theta_c=\frac{1}{\mu}\)

According to the question

⇒ \(\frac{1}{\sin \theta_c}=\frac{1}{\sin 45^{\circ}}=\frac{1}{\frac{1}{\sqrt{2}}}=\sqrt{2}\)

Velocity of light in air or vacuum,

c = 3 x 10^{8} m/s

Velocity of light in the medium

ν = \(\frac{c}{\mu}=\frac{3 \times 10^8}{\sqrt{2}}\) ‘

= 2.12

**Question 7. The refracting angle of a prism is 60° and the refractive index of its material is \(\sqrt{\frac{7}{3}}\) Find the minimum angle of incidence ofa ray of light falling on one refractingface of the prism such that the emerging ray will graze the other refracting face**

**Answer:**

⇒ \(i_1=\sin ^{-1}\left[\sin A \sqrt{\mu^2-1}-\cos A\right]\)

Here, \(\mu=\sqrt{\frac{7}{3}}\)

Or, \(\mu^2-1=\frac{4}{3}\)

⇒ \(\sqrt{\mu^2-1}=\sqrt{\frac{4}{3}}=\frac{2}{\sqrt{3}}\)

Given that A = 60

Sin A = \(\frac{\sqrt{3}}{2}\)

Cos A = \(\frac{1}{2}\)

Hence, \(\sin ^{-1} \frac{\sqrt{3}}{2} \times \frac{2}{\sqrt{3}}-\frac{1}{2}\)

= \(\sin ^{-1} \frac{1}{2}\)

= 30°

**Question 8. For the same value of angle of incidence, the angle of refraction in three media A, B and C are 15°, 25’ and 35° respectively. In which medium would the velocity of light be minimum?**

Refarctive index μ = \(\frac{\sin i}{\sin r}\)

For the same value of i, μ ∝ \(\frac{1}{\sin r}\)

In the given problrm

r_{1}<r_{2}<r_{3} Or, sin r_{1}<sinr_{2}< sinr_{3}

As velocity oflight, v = \(\frac{c}{\mu}\) i.e \(\nu \propto \frac{1}{\mu}\) , we have ν_{1}<ν_{2}<ν_{3}

So In the Grst medium, die velocity of light Is minimum.

**Question 9. Two monochromatic rays of light are incident normally on the face AB of an isosceles right-angled prism ABC. The refractive indices ofthe glass prism for the two rays 1 and 2 are respectively13 and13. Trace the path of these rays after entering through the prism**

Both the rays enter the glass prism through the face AB without deviation

Now, the critical angles for glass-to-air refraction are

For ray 1 : = \(\sin ^{-1} \frac{1}{1.3}=50.3^{\circ}\) = 50.3

For ray 1 : = \(\sin ^{-1} \frac{1}{1.5}\) = 41.8

Both the rays are incident on the surface AC at an angle of 45°. As 45° < θ_{1}, ray 1 refracted across AC to reenter air. But, as 45° > θ_{12}, ray 2 suffers total internal reflection at AC and the reflected ray is incident normally on BC and then escapes to air without deviation

**Question 10. For the same angle of incidence, the angle of refraction in two media A and B are 25° and 35° respectively. In which medium is the speed oflight less?**

We know,

∴ VA<VB

Hence, the speed of light is less in medium A

**Question 11. A ray of light incident on an equilateral glass prism propagates parallel to the base Ifne of the prism Inside it Find the angle of incidence of this ray. Given refractive index ofmaterial ofglass prism is \(\sqrt{3}\). **

From the diagram r = 30

Also \(\mu=\frac{\sin i}{\sin r}\)

Or, \(\sqrt{3}=\frac{\sin i}{\sin 30^{\circ}}\)

Or, sin i \(\sqrt{3} \times \frac{1}{2}\)

Hence i = 60

The required angle of incidence is 60°

**Question 12. A ray PQ incident on the refracting face BA is refracted in the prism BAC asshown in the figureand emerges from the other refracting face AC as BS such that AQ = AB. If the angle of prism A = 60° and refractive index of material of prism is \(\sqrt{3}\), calculate angle θ**

The angle of the prism is A = 60°. It is also given that AQ = AR. Therefore, the angles opposite to these two sides are also equal

Now for the triangle AQR

∠A+∠AQR+∠ARQ= 180

∠AQR= ∠ARQ= 60

r_{1}= r_{2}= 30

r_{1}+r_{2}= 60

When r_{1} and r_{2} are equal, we have i = e

Now, according to Snell’s law, \(\mu=\frac{\sin i}{\sin r_1}\)

\(\sin i=\mu \sin r_1=\sqrt{3} \sin 30^{\circ}=\frac{\sqrt{3}}{2}\)i = 60°

Now , the angle of deviationn

= i + e-A = 60°+60°- 60 = 60°

**Question 13. Monochromatic light of wavelength 589 nm is incident from air on a water surface. If μ for water is 1.33, find the face wavelength, frequency and speed of the refracted light
Answer:**

Velocity of light in air is nearly equal to the velocity of light in free space.

Velocity oflight in air \(\) = 2.99 × 10^{8}m. s-1

Frequency n = \(\frac{v_a}{\lambda_a}=\frac{2.99 \times 10^8}{589 \times 10^{-9}}\)

= \(5.07 \times 10^{14}\) Hz

Frequency is a fundamental property and does not change with change ofmedium

= \(\frac{\text { velocity of light in air }\left(v_a\right)}{\text { velocity of light in water }\left(v_w\right)}=\frac{n \lambda_a}{n \lambda_w}\)

= \(\frac{\lambda_a}{\mu}=\frac{589}{1.33}\)

⇒ \(frac{589}{1.33}\) = 443nm

∴ Wavelength ofthe refracted light = 443nm.

∴ Speed ofthe refracted light, vw \(=\frac{\nu_a}{\mu}=\frac{2.99 \times 10^8}{1.33}\)

How does the angle ofminimum deviation ofa glass prism vary, ifthe incident violet light is replaced by red light? give reason

We know, \(\lambda_{\text {red }}>\lambda_{\text {violet }} \text { and } \mu=A+\frac{B}{\lambda^2}\)

Where A and B are constant

So, for the increase of wavelength refractive index of the material for different coloured ray decreases

∴ \(\mu_{\text {red }}<\mu_{\text {violet }}\)

Since, \(\mu=\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \frac{A}{2}}\)

∴ \(\left(\delta_m\right)_{\text {red }}<\left(\delta_m\right)_{\text {violet }}\)

**Question 14. **

**1. A ray of light incident on face AB of an equilateral glass prism, shows minimum deviation of 30°. Calculate the speed of light through the prism**

**2. Find the angle of incidence at Find the angle of incidence at face AC.**

**Answer:**

1 . The refractive index for the material ofthe prism,

μ = \(\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \frac{A}{2}}=\frac{\sin \left(\frac{60^{\circ}+30^{\circ}}{2}\right)}{\sin 30^{\circ}}\)

[Given A = 60, = 30]

= \(\frac{\sin 45^{\circ}}{\sin 30^{\circ}}=\frac{1}{\sqrt{2}} \times 2\)

Since,

⇒ \(\mu=\frac{c}{v}\)

Speed of light through the prism = 2.12 ×10^{8}m .s^{-1}

2. Critical Angle = \(=\sin ^{-1}\left(\frac{1}{\mu}\right)\) = r_{2}

θ_{c}= \(\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)\) = 45°

A= r_{1}+ r_{2}

rx= A- r_{2}= 60°- 45° = 15°

Considering refraction at face AB ,

μ = \(\frac{\sin i_1}{\sin r_1}\)

= \(\sin i_1=\mu \sin r_1=\sqrt{2} \times \sin 15^{\circ}\)

Or, i_{1} = 21.47

**Question 15. A ray of light passing from air through an equilateral gla?s prism undergoes minimum deviation when the angle of incidence| of the angle of prifun. Calculate the speed of light in the prism.**

Here, refracting angle (A) = 60°

We know, minimum deviation (8m) = 2i_{1}– A

⇒ \(\delta_m=2 \times \frac{3}{4} A-A\)

Given \(i_1=\frac{3}{4} A\)

= \(\frac{A}{2}\)= 30

= \(\frac{A}{2}\)

Refractive index of the prism

μ = \(\frac{\sin \frac{A+\delta_m}{2}}{\sin \frac{A}{2}}=\frac{\sin 45^{\circ}}{\sin 30^{\circ}}\)

= \(\frac{1}{\sqrt{2}} \times 2\)

= \(\sqrt{2}\)

Also μ = \(\frac{c}{v}\) [ v = velocity of light in the prism]

v = \(\frac{c}{\mu}\) = ~998 x 108 = 2.12 × 10^{8} m .s^{–}^{1}

**Question 16. The figure shows a ray of light falling normally on the face AB of an equilateral glass prism having refractive index \(\frac{3}{2}\) placed in water ofrefractive index \(\frac{4}{3}\) Will this ray suffer total internal reflection on striking the face AC? Justify your answer.**

For an equilateral prism, angle of the prism, A – 60°.

Critical angle of the glass relative to water medium

⇒ \(\theta_c=\sin ^{-1}\left(\frac{1}{w^g}\right)=\sin ^{-1}\left(\frac{\mu_w}{\mu_g}\right)\)

= \(\sin ^{-1}\left(\frac{4}{3} \times \frac{2}{3}\right)=\sin ^{-1}\left(\frac{8}{9}\right)\)

On the face AC, the ray is incident at an angle of 60° which is less than 6 c. Thus the ray suffers no total internal reflection rather it is refracted in a water medium.