A ray of light in air falls on the flat surface of a glass block of refractive index \(\sqrt{2}\). The angle of incidence is \(45^\circ\). The angle of deviation of the ray as it enters the glass is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(15^\circ\)
The ray bends towards the normal as it enters the glass. Its deviation at this surface is the angle between the incident ray and the refracted ray, \( \delta=i-r \).
\( i=45^\circ \)
\( n=\sqrt{2} \)
\( \begin{aligned} \sin r &= \frac{\sin i}{n} \\ &= \frac{1/\sqrt{2}}{\sqrt{2}} \\ &= \frac{1}{2} \end{aligned} \)
\( r=30^\circ \)
\( \begin{aligned} \delta &= i-r \\ &= 45^\circ-30^\circ \\ &= 15^\circ \end{aligned} \)
Hence, the ray turns through \(15^\circ\) when it enters the glass.
NCERT (2026-27):
Refraction – “(ii) The ratio of the sine of the angle of incidence to the sine of angle of refraction is constant. … We have”
\( \displaystyle \frac{\sin i}{\sin r}=n_{21} \)\( \qquad (9.10) \)
“where n21 is a constant, called the refractive index of the second medium with respect to the first medium.”
Refraction Through a Prism – “… The total deviation δ is the sum of deviations at the two faces, δ= (i – r1 ) + (e – r2 )”