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

Ans (10.00) μ 1 sin θ 1 = μ 2 sin θ 2 3 5 sin 3 0 ∘ = 3 4 sin θ 2 ∴ θ 2 = sin − 1 ( 8 5 ) for total internal reflection at P, we use 3 5 sin 6 0 ∘ 2 3 5

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 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:

Ans (10.00)

$μ_{1} sin θ_{1} = μ_{2} sin θ_{2}$

$\frac{5}{3} sin 3 0^{\circ} = \frac{4}{3} sin θ_{2} ∴ θ_{2} = sin^{- 1} (\frac{5}{8})$

for total internal reflection at P, we use $\frac{5}{3} sin 6 0^{\circ} < n$

$n > \frac{5}{2 3}$

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

⏱ 2mℹ️ Source: QB

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