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Question

Physics Question on Refraction of Light

For the angle of minimum deviation of a prism to be equal to its refracting angle, the prism must be made of a material whose refractive index :

A

lies between 2\sqrt {2} and 11

B

lies between 2 and 2\sqrt {2}

C

is less then 1

D

is greater then 2

Answer

lies between 2 and 2\sqrt {2}

Explanation

Solution

The angle of minimum deviation is given as
δmin=i+eA\delta_{\min }=i+e-A
for minimum deviation
δmin=A\delta_{\min }= A then
2A=i+e2 A = i + e
in case of δmin;i=e\delta_{\min } ; i=e
2A=2ir1=r2=A22 A =2 i r _{1}= r _{2}=\frac{ A }{2}
i=A=90i = A =90^{\circ}
1sinr=nsinr11 \sin r = n \sin r _{1}
sinA=nsinA2\sin A = n \sin \frac{ A }{2}
2sinA2cosA2=nsinA22 \sin \frac{ A }{2} \cos \frac{ A }{2}= n \sin \frac{ A }{2}
2cosA2=n2 \cos \frac{ A }{2}= n
when A=90=iminA=90^{\circ}=i_{\min }
then nmin=2n_{\min }=\sqrt{2}
i=A=0i=A=0
nmax=2n_{\max }=2