JEE MAIN - Physics (2025 - 28th January Morning Shift - No. 16)
A thin prism $\mathrm{P}_1$ with angle $4^{\circ}$ made of glass having refractive index 1.54 , is combined with another thin prism $\mathrm{P}_2$ made of glass having refractive index 1.72 to get dispersion without deviation. The angle of the prism $\mathrm{P}_2$ in degrees is
1.5
16/3
3
4
Explanation
We know, for thin prism,
$$\mu = {{A + \delta } \over A}$$
where, $\mu$ = refractive index
A = Angle of prism
$\delta$ = Angle of deviation
We can write, $$\mu A = A + \delta $$
$$ \Rightarrow \delta = (\mu - 1)A$$
given, $${\delta _{net}} = 0$$
$$ \Rightarrow ({\mu _1} - 1){A_1} - ({\mu _2} - 1){A_2} = 0$$
$$ \Rightarrow (1.54 - 1)4 - (1.72 - 1){A_2} = 0$$
$$ \Rightarrow {A_2} = {{54 \times 4} \over {72}} \Rightarrow {A_2} = 3^\circ $$
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