JEE Main Physics Class 12 Ray Optics and Optical Instruments Part 1 Questions

When r2=C,∠N2RC=90°
Where C= critical angle
As sin‌C=

1

µ

=sin‌r2
Applying snell's law at 'R'
µ‌sin‌r2=1‌sin‌90° ......(i)
Applying snell's law at 'Q'
1×sin‌θ=µ‌sin‌r1 ......(ii)
But r1=A−r2
So, sin‌θ=µ‌sin(A−r2)
sin‌θ=µ‌sin‌A‌cos‌r2−cos‌A .......(iii) [using (i)]
From (1)
cos‌r2=√1−sin2r2=√1−

1

µ2

.......(iv)
By eq. (iii) and (iv)
sin‌θ=µ‌sin‌A‌√1−

1

µ2

−cos‌A
on further solving we can show for ray not to transmitted through face AC
θ=sin−1[µ‌sin(A−sin−1(

1

µ

)]
So, for transmission through face AC
θ>sin−1[µ‌sin(A−sin−1(

1

µ

)]