The exit angle θ 2 of the ray is sin − 1 ( 16 x ) , find the value of x .

3 5 sin 3 0 ∘ = 3 4 sin θ 2 , so sin θ 2 = 8 5 16 x = 8 5 , so x = 10 Answer 10.00

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Physics - Geometrical Optics Question with Solution | TestHub

PhysicsGeometrical Optics200Medium2 minQB

PhysicsMediumnumerical

Passage / Comprehension

A ray of light is incident normally on one face of $3 0^{\circ} - 6 0^{\circ} - 9 0^{\circ}$ prism of refractive index $μ_{P} = 5/3$ immersed in water of refractive index $μ_{m} = 4/3$ as shown in the figure.\\

The exit angle $θ_{2}$ of the ray is $sin^{- 1} (\frac{x}{16})$ , find the value of x .

Answer:

10.00

Solution:

$\frac{5}{3} sin 3 0^{\circ} = \frac{4}{3} sin θ_{2}$ , so $sin θ_{2} = \frac{5}{8}$

$\frac{x}{16} = \frac{5}{8}$ , so x = 10

Answer 10.00

Stream:JEESubject:PhysicsTopic:Geometrical OpticsSubtopic:200

⏱ 2mℹ️ Source: QB

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