A ray is incident from glass on the boundary separating glass and water- Refractive index for glass is 32 and refractive index for water is 43- The critical angle for the glass-water boundary is

Given: Refractive index of glass (Denser medium), μ_d=32 Refractive index of water (Rarer medium), μ_r=43 From formula, i_c=sin^ 1(μ_rμ_d) ⇒ i_c=sin^ 1(4/33/2) ...

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Standard XII

Physics

Mirror Formula Formal Proof

A ray is inci...

Question

A ray is incident from glass on the boundary separating glass and water. Refractive index for glass is $\frac{3}{2}$ and refractive index for water is $\frac{4}{3}$ . The critical angle for the glass-water boundary is

{sin}^{- 1} (\frac{2}{3})

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{sin}^{- 1} (\frac{8}{9})

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{sin}^{- 1} (\frac{3}{4})

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{sin}^{- 1} (\frac{1}{9})

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Solution

The correct option is B ${sin}^{- 1} (\frac{8}{9})$
Given:
Refractive index of glass (Denser medium), $μ_{d} = \frac{3}{2}$
Refractive index of water (Rarer medium), $μ_{r} = \frac{4}{3}$
From formula,
$i_{c} = {sin}^{- 1} (\frac{μ_{r}}{μ_{d}})$
$\Rightarrow i_{c} = {sin}^{- 1} (\frac{4 / 3}{3 / 2})$
$\Rightarrow i_{c} = {sin}^{- 1} (\frac{8}{9})$

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Mirror Formula Formal Proof

Standard XII Physics

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A ray is incident from glass on the boundary separating glass and water. Refractive index for glass is 32 and refractive index for water is 43. The critical angle for the glass-water boundary is

The correct option is B sin−1(89)Given: Refractive index of glass (Denser medium), μd=32 Refractive index of water (Rarer medium), μr=43 From formula, ic=sin−1(μrμd) ⇒ic=sin−1(4/33/2) ⇒ic=sin−1(89)

A ray is incident from glass on the boundary separating glass and water. Refractive index for glass is $\frac{3}{2}$ and refractive index for water is $\frac{4}{3}$ . The critical angle for the glass-water boundary is