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Three prisms 1,2 and 3 have the prism angle $A=60^{\circ}$, but their refractive indices are respectively 1.4, 1.5 and 1.6. If $\delta_1, \delta_2, \delta_3$ be their respective angles of deviation then
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Verified Answer
The correct answer is:
$\delta_3\gt\delta_2\gt\delta_1$
$\delta=(\mu-1) \mathrm{A}($ for small $\mathrm{A})$
So, $\delta_1=\left(\mu_1-1\right) \mathrm{A}$
$=(1.4-1) \mathrm{A}$
$=0.4 \mathrm{~A}$
$\delta_2=\left(\mu_2-1\right) \mathrm{A}$
$=(1.5-1) \mathrm{A}$
$=0.5 \mathrm{~A}$
$\delta_3=\left(\mu_3-1\right) \mathrm{A}$
$=(1.6-1) \mathrm{A}$
$=0.6 \mathrm{~A}$
Clearly, $\delta_3\gt\delta_2\gt\delta_1$
So, $\delta_1=\left(\mu_1-1\right) \mathrm{A}$
$=(1.4-1) \mathrm{A}$
$=0.4 \mathrm{~A}$
$\delta_2=\left(\mu_2-1\right) \mathrm{A}$
$=(1.5-1) \mathrm{A}$
$=0.5 \mathrm{~A}$
$\delta_3=\left(\mu_3-1\right) \mathrm{A}$
$=(1.6-1) \mathrm{A}$
$=0.6 \mathrm{~A}$
Clearly, $\delta_3\gt\delta_2\gt\delta_1$
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