Question

Consider An equilateral prism made of silica glass with n=1.5. Light is incident on one face at an angle of π/4 with respect to the normal to thatface. Determine the angle at which the light emerges from the opposite face of the prism with respect to the original incident beam direction.

Answer 1

Answer:

The angle at which the light emerges from the opposite face of the prism with respect to the original incident beam is 53.3°.

Explanation:

Given that,

Angle $\theta=\dfrac{\pi}{4}$

$\theta=45$

Refractive index n=1.5

For the refraction at the first surface

We need to calculate the angle

Using formula of refraction

$n_{a}\sin\theta_{1}=n_{g}\sin\theta_{2}$

Put the value into the formula

$1\sin\dfrac{\pi}{4}=1.5\sin\theta_{2}$

$\sin\theta_{2}=\dfrac{1\times\sin45}{1.5}$

$\sin\theta_{2}=0.46520$

$\theta_{2}=\sin^{-1}(0.46520)$

$\theta_{2}=27.7^{\circ}$

We need to calculate the angle on incidence at the second surface

$(90-\theta_{2})+(90-\theta_{3})+A=180$

$\theta_{3}=A-\theta_{2}$

Put the value into the formula

$\theta_{3}=60-27.7$

$\theta_{3}=32.3$

For the refraction at the second surface,

We need to calculate the angle at which the light emerges from the opposite face of the prism with respect to the original incident beam direction

Using formula of refraction

$n_{g}\sin\theta_{3}=n_{a}\sin\theta_{4}$

Put the value into the formula

$1.5\sin32.3=1\times\sin\theta_{4}$

$\theta=\sin^{-1}(\dfrac{1.5\sin32.3}{1})$

$\theta=53.3^{\circ}$

The angle is 53.3° from the original

Hence, The angle at which the light emerges from the opposite face of the prism with respect to the original incident beam is 53.3°.

Answer 2

Answer:

$r'=36.0113^{\circ}$

Explanation:

Given:

• Equilateral prism of silica glass.

• refractive index of the silica glass, $n=1.5$

• angle of incidence on the air-glass interface, $\angle i=\frac{\pi}{4} =45^{\circ}$

Form the Snell's law we know:

$n=\frac{sin\ i}{sin\ r}$ ...................(1)

where:

$r=$ angle of refraction when light passes from air to glass.

putting values in (1)

$1.5=\frac{sin\ 45^{\circ}}{sin\ r}$

$r= 28.125^{\circ}$

Now refer the schematic for further geometric relations involved:

From the triangle LMN:

$r+i'=90^{\circ}$

$i'=61.875^{\circ}$ is the second angle of incidence when the light ray incidents on glass-air interface when passing from glass to air.

Again using Snell's law:

$1.5=\frac{sin\ 61.875^{\circ}}{sin\ r'}$

$r'=36.0113^{\circ}$ which is the angle of emergence.