Refraction, TIR and Prism — NEET Physics PYQ MCQs with Solutions

Solution: Critical angle $i_C = 45°$ $\mu = \frac{1}{\sin i_C} = \frac{1}{\sin 45°} = \sqrt{2}$ $\frac{c}{v_m} = \sqrt{2} \Rightarrow \frac{3 \times 10^8}{v_m} = \sqrt{2}$ $\Rightarrow v_m = \frac{3}{\sqrt{2}} \times 10^8$ m/s

Solution: Since the ray emerges normally from the second face, the angle of refraction at that face $r_2 = 0$. As $A = r_1 + r_2$, we get $r_1 = A$. From Snell's law, $\sin i = \mu \sin A$ For small angles, $\sin\theta \approx \theta$, so $i = \mu A$.

Solution: The necessary conditions for a rainbow are: (i) The sun should be shining in one part of the sky while it is raining in the opposite part. (ii) The observer must stand with his back towards the sun. So the statement that an observer can see a rainbow when his front is towards the sun is wrong; the rest are correct.

Solution: Different colours of white light have different wavelengths. In descending order: $\lambda_{Red} > \lambda_{Green} > \lambda_{Blue} > \lambda_{Violet}$ Red has the longest wavelength.

Solution: The critical angle for a pair of media in contact is defined as the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is 90°.

Solution: For the ray to retrace its path, it must fall normally on the silvered surface. Geometry of the prism then gives the angle of refraction at the first face $r_1 = 30°$. Applying Snell's law at the first face: $\mu = \frac{\sin i}{\sin r_1} \Rightarrow \sqrt{2} = \frac{\sin i}{\sin 30°}$ $\sin i = \sqrt{2} \times \frac{1}{2} = \frac{1}{\sqrt{2}} \Rightarrow i = 45°$

Solution: For dispersion without deviation, the net deviation produced by the combination must be zero: $(\mu_1 - 1)A_1 = (\mu_2 - 1)A_2$ $A_2 = \left(\frac{\mu_1 - 1}{\mu_2 - 1}\right)A_1 = \left(\frac{1.42 - 1}{1.7 - 1}\right)(10°) = 6°$

Solution: At minimum deviation, $i = e = 45°$ and $r = r' = \frac{A}{2} = 30°$. $\delta_{min} = i + e - (r + r') = 90° - 60° = 30°$ $\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\frac{A}{2}} = \frac{\sin 45°}{\sin 30°} = \frac{1/\sqrt{2}}{1/2} = \sqrt{2}$

Solution: Let the thickness of the slab be t and the bubble be at depth x from one face. Apparent depths from the two faces: $\frac{x}{\mu} + \frac{t - x}{\mu} = 3 + 5$ $\frac{t}{\mu} = 8$ cm $t = 8\mu = 8 \times \frac{3}{2} = 12$ cm

Solution: $\mu = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\frac{A}{2}}$ Given $\mu = \cot\frac{A}{2} = \frac{\cos(A/2)}{\sin(A/2)}$, so $\sin\left(\frac{A + D_m}{2}\right) = \cos\frac{A}{2} = \sin\left(\frac{\pi}{2} - \frac{A}{2}\right)$ $\frac{A + D_m}{2} = \frac{\pi}{2} - \frac{A}{2}$ $D_m = \pi - 2A = 180° - 2A$

Solution: At the hypotenuse the light strikes at 45°. Total internal reflection occurs if the angle of incidence exceeds the critical angle, i.e. if $\mu > \frac{1}{\sin 45°} = \sqrt{2} \approx 1.414$ Since $\mu_{red} = 1.39 < 1.414$, red light refracts out; green (1.44) and blue (1.47) undergo total internal reflection. Thus only the red colour is separated from green and blue.

Solution: Since the ray retraces its path, it strikes the silvered surface normally. Geometry then gives the angle of refraction at the first face $r = A$. By Snell's law: $\mu = \frac{\sin 2A}{\sin A} = \frac{2\sin A\cos A}{\sin A} = 2\cos A$

Solution: Since the ray emerges normally, $r_2 = 0$, so $r_1 = A$. $\mu = \frac{\sin i}{\sin r_1} = \frac{\sin i}{\sin A}$ For small angles $\sin i \approx i$ and $\sin A \approx A$: $\mu = \frac{i}{A} \Rightarrow i = \mu A$

Solution: Real and apparent depth are explained on the basis of refraction only — the concept of total internal reflection is not involved. Mirage, brilliance of diamond and optical fibres all involve total internal reflection.

Solution: For total internal reflection, $i > i_C$, i.e. $\sin i > \sin i_C$ $\sin 45° > \frac{1}{\mu}$ $\mu > \sqrt{2} \approx 1.414$ Only $\mu = 1.50$ satisfies this.

Solution: Velocity of light in the material: $v = \nu\lambda$ Refractive index: $\mu = \frac{c}{v} = \frac{c}{\nu\lambda}$ $\mu = \frac{3 \times 10^8}{2 \times 10^{14} \times 5000 \times 10^{-10}} = 3.00$

Solution: Since the ray grazes along the surface, the angle of incidence equals the critical angle C. $\tan C = \frac{R}{h} = \frac{3}{4}$, so $\sin C = \frac{3}{5}$ $\mu = \frac{1}{\sin C} = \frac{5}{3}$ $v = \frac{c}{\mu} = \frac{3 \times 10^8}{5/3} = 1.8 \times 10^8$ m/s

Solution: For the beam to retrace its path it must strike the mirrored face normally. The geometry then gives the angle of refraction at the entry face $r = 30°$. By Snell's law: $\mu = \frac{\sin i}{\sin r} \Rightarrow \sqrt{2} = \frac{\sin i}{\sin 30°}$ $\sin i = \sqrt{2} \times \frac{1}{2} = \frac{1}{\sqrt{2}} \Rightarrow i = 45°$

Solution: In any medium other than air or vacuum, the velocities of different colours are different, so red and green are refracted at different angles inside the slab. After emerging through the opposite parallel face, they appear at two different points and travel in two different parallel directions (each parallel to the incident beam).

Solution: At the first face: $\frac{\sin 45°}{\sin r} = \mu \Rightarrow \sin r = \frac{1}{\sqrt{2}\mu}$ At the adjacent face the angle of incidence is $\theta = 90° - r$. For TIR: $\sin(90° - r) > \frac{1}{\mu}$, i.e. $\cos r > \frac{1}{\mu}$ $\cos r = \sqrt{1 - \frac{1}{2\mu^2}}$, so $1 - \frac{1}{2\mu^2} \geq \frac{1}{\mu^2} \Rightarrow \mu^2 \geq \frac{3}{2}$ $\mu = \sqrt{\frac{3}{2}}$

Solution: An optical fibre is a device based on total internal reflection by which a light signal is transferred from one place to another with negligible loss of energy. Light incident at one end at a small angle suffers multiple total internal reflections along the fibre and emerges at the other end.

Solution: $\mu = \frac{\text{real depth}}{\text{apparent depth}}$ Net apparent depth $= 2 + 5 = 7$ cm Real depth (edge of cube) $=$ apparent depth $\times \mu = 7 \times 1.5 = 10.5$ cm

Solution: When white sunlight falls on raindrops, the small drops of water behave like prisms. Refraction, dispersion and total internal reflection of white light occur inside the water drops, producing the band of colours seen as a rainbow.

Solution: $\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\frac{A}{2}}$, with $A = 60°$ $\sqrt{3} = \frac{\sin\left(\frac{60° + \delta_m}{2}\right)}{\sin 30°}$ $\sin\left(\frac{60° + \delta_m}{2}\right) = \frac{\sqrt{3}}{2} = \sin 60°$ $\frac{60° + \delta_m}{2} = 60° \Rightarrow \delta_m = 60°$

Solution: At the lateral face the angle of incidence is $\theta = 90° - r$. For no refraction there (TIR): $\cos r > \sin C = \frac{1}{\mu}$ With $\sin r = \frac{\sin i}{\mu}$ this gives $1 - \frac{\sin^2 i}{\mu^2} > \frac{1}{\mu^2}$, i.e. $\mu^2 > \sin^2 i + 1$ The maximum value of $\sin i$ is 1, so $\mu^2 > 2$, i.e. $\mu > \sqrt{2}$

Solution: When an electromagnetic wave enters another medium, its frequency remains unchanged while both wavelength and velocity become $\frac{1}{\mu}$ times. So in the glass slab: frequency $\nu$, wavelength $\frac{\lambda}{\mu}$, velocity $\frac{v}{\mu}$.

Solution: By Cauchy's relation $\mu \propto \frac{1}{\lambda}$, and from the lens maker's formula $f \propto \frac{1}{\mu - 1}$, hence $f \propto \lambda$. For a convex lens: $f_V < f_R$. For a concave lens the focal length is negative, so $F_V > F_R$ (violet focal length is less negative).

Solution: Let y be the apparent position of the silvered surface (acting as a plane mirror). For a plane mirror, object distance = image distance: $y + 8 = 12 + 6 - y$ $y = 5$ cm $\mu = \frac{\text{real depth}}{\text{apparent depth}} = \frac{6}{5} = 1.2$

Solution: Speed of light in the glass plate $= \frac{c}{\mu}$ Time taken $= \frac{t}{c/\mu} = \frac{\mu t}{c}$

Solution: When two or more lenses are combined so that the combination is free from chromatic aberration, it is called an achromatic combination. For this, one lens should be convex and the other concave.