Question

monochromatic Ray of light is incident at 37° on an equilateral prism of refractive index 3/2. Determine angle of emergence and angle of deviation. If angle of prism is adjustable, what should it's value be for emergent ray to be just possible for the same angle of incidence.​

Answer 1

Answer:

Angle of emergence is 62.9 degree

deviation angle is 39.9 degree

if angle of emergence and angle of incidence is same then angle of prism will be 47.2 degree

Explanation:

Angle of incidence of the light on the surface of prism is given as

$\theta = 37 degree$

now by Snell's law

$1 sin 37 = \frac{3}{2} sin\theta$

$\frac{3}{5} = \frac{3}{2} sin\theta$

$\theta = 23.6 degree$

now on the other surface of the prism the angle of incidence is given as

$\theta + \phi = A$

$23.6 + \phi = 60$

$\phi = 36.4$degree

now we have

$\frac{3}{2}sin36.4 = 1 sin\alpha$

$\frac{3}{2}(0.59) = sin\alpha$

$\alpha = 62.9$

now we have

$\delta = i + e - A$

$\delta = 37 + 62.9 - 60$

$\delta = 39.9 degree$

Part b)

if angle of incidence = angle of emergence

so we have

$r_1 = r_2 = 23.6$

$A = r_1 + r_2$

$A = 47.2 degree$

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Topic : Prism

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