Question

A prism is made of glass of unknown refractive index. A parallel
beam of light is incident on a face of the prism. The angle of minimum
deviation is measured to be 40°. What is the refractive index of the
material of the prism? The refracting angle of the prism is 60°. If
the prism is placed in water (refractive index 1.33), predict the new
angle of minimum deviation of a parallel beam of light.



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Answer 1

Answer:

1.532, 29.661 °

Explanation:

When the minimum angle of deviation and the refracting angle of a prism is given, use the formula: refractive index, u= sin{(A+ dev.min.)/2} /sin(A/2)

Look at the attachment for better explanation.

Excuse me if I have made calculation error.

Answer 2

$\large {\bold{ \underline{ Given \: ,}}}$

• $\small \sf{Angle \: of \: minimum \: deviation \: , \: δ_{m}=40 {\degree}}$

• $\small \sf{Angle \: of \: the \: prism, A=60 {\degree}}$

• $\small \sf{Refractive \: index \: of \: water, μ_{w} =1.33.}$

$\bold {Let \: the \: refractive \: index \: of \: the \: material \: of \: the \: prism \: be \: μ_{g}}$

$\bold {\underline{Using \: the \: formula, }} \: { \bigstar}$

$: \implies \tt {\red{μ_{g} = \dfrac{Sin (+δ_{m}/2 )}{Sin( \dfrac{A }{2})}}}$

$\bold{we \: have,}$

$: \implies \tt{μ _{g} = Sin( \dfrac{Sin </p><p> \dfrac{(60+40)}{2}}{ \dfrac{60}{2}} )}$

$: \implies \tt{μ _</p><p>{g} = \dfrac{Sin50}{Sin30} }$

$: \implies \tt \red{μ _</p><p>{g}=1.532}$

• Thus, the refractive index of the material of the prism is 1.532.

Since the prism is placed in water, let δₘw be the new angle of minimum deviation for the same prism.

The refractive index of glass with respect to water is given by the relation:

$: \implies \tt</p><p>w μ _{g} = \dfrac{Sin \dfrac{(A+δ _{m }w)}{2} }{Sin( \dfrac{A}{2} )}$

$: \implies \tt{wμ _{g} = \dfrac{μ_{g} }{μ _{w}} = \dfrac{Sin \dfrac{(A+δ _{m }w)}{2}}{Sin( \dfrac{A}{2} ) }}$

$: \implies \tt{w μ _{g}= \dfrac{1.532}{1.33} = \dfrac{Sin \dfrac{(60+δ _{m }w )}{2}}{Sin( \dfrac{60}{2}) }}$

$: \implies \tt{wμ _{g} = \dfrac{1.532}{</p><p>1.33} \times Sin( \dfrac{60}{2} )= Sin \dfrac{(60+δ _{m }w )}{2}}$

$: \implies \tt{w μ _{g} = 0.575 = Sin \dfrac{(60+δ _{m }w )}{2}}$

$: \implies \tt{ \sin^{- 1}(0.575) = \dfrac{(60+δ _{m }w)}{2}}$

$: \implies\tt{60+δ_{m}w=35.16×2δ _{m }w}$

$: \implies \tt{ 70.33−60δ _{m }w }$

$: \implies \tt {\red{ 10.33 \degree</p><p> }}\:{\bigstar}$

• Thus, new angle of minimum deviation is 10.33∘