Question

For an isosceles prism of angle $

• A. For the angle of incidence $
• B. For this prism, the refractive index $\mu$ and the angle of prism
• C. At minimum deviation, the incident angle $
• D. For this prism, the emergent ray at the second surface will be tangential to the surface when the angle of incidence at the first surface is $i_{1}=\sin^{-1}\left[\sin\,A\sqrt{\frac{4\,\cos^{2}}{A}2 -1-\cos\,A}\right]$

Answer 1

Answer: (D)

For this prism, the emergent ray at the second surface will be tangential to the surface when the angle of incidence at the first surface is $i_{1}=\sin^{-1}\left[\sin\,A\sqrt{\frac{4\,\cos^{2}}{A}2 -1-\cos\,A}\right]$

Answer 2

(A) At minimum deviation
$i = e$
$\delta_{\min }=2 i - A$
$i =\frac{\delta_{ \min }+ A }{2}= A$ ...(I)
$r _{1}= r _{2}=\frac{ A }{2}$ ...(II)
$r _{1}=\frac{ i }{2}$
(B) $\mu=\frac{\sin \left(\frac{\delta_{ m }+ A }{2}\right)}{\sin \left(\frac{ A }{2}\right)}=\frac{\sin A }{\sin \frac{ A }{2}}=2 \cos \frac{ A }{2}$
$A =2 \cos ^{-1}\left(\frac{\mu}{2}\right)$
(C) If $i _{1}= A$, deviation is minimum, ray is parallel to the base.
(D) When $e =90^{\circ}$
$r _{2}=\theta_{c}$
$r _{1}= A - r _{2}= A -\theta_{ c }$
$\sin i =\mu \sin r _{1}$
$\sin i =\mu \sin \left( A -\theta_{c}\right)$
$\sin i =\mu\left(\sin A \cos \theta_{c}-\cos A \sin \theta_{c}\right)$
$\sin i =\mu\left[\sin A \sqrt{1-\sin ^{2} \theta_{ c }}-\cos A \cdot \sin \theta_{c}\right]$
$\sin i =\mu\left[\sin A \sqrt{1-\frac{1}{\mu^{2}}}-\cos A \cdot \frac{1}{\mu}\right]$
$\sin i =\mu \times \frac{1}{\mu}\left[\sin A \sqrt{\mu^{2}-1}-\cos A \right]$
$\sin i =\left[\sin A \sqrt{4 \cos ^{2} \frac{ A }{2}-1}-\cos A \right]$
$i =\sin ^{-1}\left[\sin A \sqrt{4 \cos ^{2} \frac{ A }{2}-1}-\cos A \right]$