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

Normal entry on the first face: no bending; incidence at the exit face is 3 0 ∘ . μ P sin θ 1 = μ m sin θ 2 : 3 5 sin 3 0 ∘ = 3 4 sin θ 2 . sin θ 2 = 8 5 = 16 10 x = 10.00

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Physics - Geometrical Optics Question with Solution | TestHub

PhysicsGeometrical OpticsTIR/Prism and DispersionMedium2 minQB

PhysicsMediumnumerical

Passage / Comprehension

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

Normal entry on the first face: no bending; incidence at the exit face is $3 0^{\circ}$ .

$μ_{P} sin θ_{1} = μ_{m} sin θ_{2}$ : $\frac{5}{3} sin 3 0^{\circ} = \frac{4}{3} sin θ_{2}$ .

$sin θ_{2} = \frac{5}{8} = \frac{10}{16}$

$x = 10.00$

Stream:JEESubject:PhysicsTopic:Geometrical OpticsSubtopic:TIR/Prism and Dispersion

⏱ 2mℹ️ Source: QB

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