A rectangular block of glass $(R . l . = 3 / 2)$ is kept in water $(R . l . = 4 / 3)$ . The critical angle for total. Internal reflection between glass and water is

{sin}^{- 1} (2 / 3)

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{sin}^{- 1} (3 / 4)

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{sin}^{- 1} (8 / 9)

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{sin}^{- 1} (1 / 2)

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Solution

The correct option is C ${sin}^{- 1} (8 / 9)$
We know,
$i_{c} = {sin}^{- 1} (\frac{μ_{r a r e r}}{μ_{d e n s e r}})$

Given,
$μ_{g l a s s} = \frac{3}{2}, μ_{w a t e r} = \frac{4}{3}$

For Glass - Water interface,

$i = {sin}^{- 1} ⎛ ⎜ ⎜ ⎜ ⎝ \frac{\frac{4}{3}}{\frac{3}{2}} ⎞ ⎟ ⎟ ⎟ ⎠ = {sin}^{- 1} (\frac{8}{9})$

Hence option $C$ is the correct answer.

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Similar questions

Q. Show that

\frac{9 π}{8} - \frac{9}{4} {sin}^{- 1} (\frac{1}{3}) = \frac{9}{4} {sin}^{- 1} (\frac{2 \sqrt{2}}{3})

Q. Critical angle of glass is

θ_{1}

and that of water is

θ_{2}

. The critical angle for water and glass surface would be

(μ_{g} = 3 / 2, μ_{w} = 4 / 3)

\frac{9 π}{8} - \frac{9}{4} {sin}^{- 1} \frac{1}{3} = \frac{9}{4} {sin}^{- 1} \frac{2 \sqrt{2}}{3}

Q. Prove:

\frac{9 π}{8} - \frac{9}{4} {sin}^{- 1} \frac{1}{3} = \frac{9}{4} {sin}^{- 1} \frac{2 \sqrt{2}}{3}

Critical angle of glass is $θ_{1}$ and that of water is $θ_{2}$ . The critical angle for water and glass surface would be:
$(μ_{g} = \frac{3}{2}, μ_{w} = \frac{4}{3})$

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A rectangular block of glass (R.l.=3/2) is kept in water (R.l.=4/3). The critical angle for total. Internal reflection between glass and water is

The correct option is C sin−1(8/9)We know,ic=sin−1(μrarerμdenser)Given,μglass=32,μwater=43For Glass - Water interface,i=sin−1⎛⎜ ⎜ ⎜⎝4332⎞⎟ ⎟ ⎟⎠=sin−1(89)Hence option C is the correct answer.

A rectangular block of glass $(R . l . = 3 / 2)$ is kept in water $(R . l . = 4 / 3)$ . The critical angle for total. Internal reflection between glass and water is