Question

A light ray is incident at an angle of 45° with the normal to a √2 cm thick plate (μ = 2.0). Find the shift in the path of the light as it emerges out from the plate.

Answer 1

Dear Student,

◆ Answer -

∆s = 0.622 cm

● Explanation -

# Given -

t = √2 cm

i = 45°

μ = 2

# Solution -

To find r,

μ = sini/sinr

2 = sin45° / sinr

2 = 0.7071 / sinr

sinr = 0.7071 / 2

sinr = 0.3536

r = 20.7°

Shift in the path of light is -

∆s = tsin(i-r) / cosr

∆s = √2 × sin(45-20.7) / cos20.7

∆s = 1.414 × sin24.3 × 0.9354

∆s = 1.414 × 0.4115 × 1.069

∆s = 0.622 cm

Hence, shift in the path of the light is 0.622 cm.

Thanks dear. Hope this helps you..

Answer 2

The shift in the path of the light as it emerges out from the plate is 0.62 cm.

Explanation:

Given data in the question  :

A light-ray happens at an angle of 45° with Standard plate with a thickness of  √2 cm (μ = 2.0).

As in the diagram,

                             $\frac{\sin 45^{\circ}}{\sin r}=\frac{2}{1}$

                               $2 \sin r=\sin 45^{\circ}$

                               $2 \sin r=\frac{1}{\sqrt{2}}$

                                  $\sin r=\frac{1}{2 \sqrt{2}}$

Therefore,  

                  θ = 45°-21° = 24°

                BD =  SHIFT IN PATH = AB sin24°

          sin 24° = 0.406

                       = AB × 0.406

                       $=\frac{\mathrm{AE}}{\cos 21^{\circ}} \times 0.406$

                       $=\frac{\mathrm{AE}}{0.933} \times 0.406$

                  AE = √2  (given in the diagram)

                       $=\frac{\sqrt{2}}{0.933} \times 0.406$

                       $=\frac{1.41}{0.933} \times 0.406$

                       = 1.515× 0.406

                       = 0.615 cm

Thus, the shift in the path of the light is 0.62 cm.