Ray Optics and Optical Instruments — NEET Physics PYQ MCQs with Solutions

Solution: The minimum size of a plane mirror required to see the full image of a person $= \frac{\text{height of person}}{2} = \frac{6}{2} = 3$ ft

Solution: When an object is held between two plane mirrors inclined at angle $\theta$, multiple reflections form several images. The total number of images is $n = \frac{360°}{\theta} - 1$ Here $\theta = 60°$, so $n = \frac{360°}{60°} - 1 = 6 - 1 = 5$

Solution: Ray optics uses the geometry of straight lines to account for macroscopic phenomena like rectilinear propagation, reflection and refraction. It is the limiting case of wave optics, valid when the dimensions involved are much larger than the wavelength of light.

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: 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: 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: 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: 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.

Solution: When the refracting angle of a prism is small (about 10° or less), the deviation is calculated from the relation $\delta = (\mu - 1)A$. For prisms with larger refracting angles we use $\delta = (i_1 + i_2) - A$.

Solution: During refraction the frequency $\nu$ remains constant while the velocity decreases in the denser medium. $\lambda_m = \frac{\lambda_v}{\mu}$, and since $\mu > 1$, $\lambda_m < \lambda_v$ Hence the wavelength decreases in the medium.

Solution: A spherometer is used to measure the radius of curvature of a spherical surface — such as the curved surface of a lens or a curved mirror.

Solution: If the biconvex lens behaves like a plane sheet of glass, rays pass undeviated through it. This happens only when the surrounding medium has the same refractive index as the lens material.

Solution: For thin lenses in contact: $\frac{1}{F_{eq}} = \frac{1}{f_1} + \frac{1}{f_2}$ $P_{eq} = \frac{1}{F_{eq}} = \frac{f_1 + f_2}{f_1 f_2}$

Solution: $\frac{1}{F} = \frac{1}{25} + \frac{1}{-25} = 0$ $F = \infty$ Power of the combination $P = \frac{1}{F} = 0$ D

Solution: Since $\lambda_{violet} < \lambda_{red}$, we have $\mu_{violet} > \mu_{red}$, and as $\frac{1}{f} \propto (\mu - 1)$: $f_{red} > f_{violet}$ The focal length is maximum for red light.

Solution: The magnification of an astronomical telescope is directly proportional to the focal length of the objective. A large aperture contributes to better quality and visibility of images, gives better resolution, and ensures better light gathering power. So all the statements are correct.

Solution: A hypermetropic (long-sighted) person can see clearly only distant objects — the near point of the eye shifts to a farther point. This defect arises due to contraction of the eyeball or increase in the focal length of the eye lens, and is corrected using a convex lens of suitable focal length.

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: 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: For the lens: $u = -60$ cm, $f = +30$ cm $\frac{1}{v} = \frac{1}{30} - \frac{1}{60} \Rightarrow v = 60$ cm This image (60 cm behind the lens, i.e. 20 cm behind the mirror) acts as a virtual object for the plane mirror, which forms a real image 20 cm in front of the mirror (20 cm from the lens). This image acts as an object for the lens again: $\frac{1}{v} + \frac{1}{20} = \frac{1}{30} \Rightarrow v = -60$ cm from the lens i.e. 20 cm from the plane mirror, and virtual in nature.

Solution: The combination is two glass lenses (focal length f each) with a bi-concave water lens between them. For the water lens: $\frac{1}{f_2} = \left(\frac{4}{3} - 1\right)\left(-\frac{2}{R}\right) = -\frac{2}{3}\cdot\frac{1}{f}$ (using $\frac{1}{f} = \frac{1}{R}$ for each glass lens) $\frac{1}{f_{eq}} = \frac{2}{f} - \frac{2}{3f} = \frac{6 - 2}{3f} = \frac{4}{3f}$ $f_{eq} = \frac{3f}{4}$

Solution: Each plano-convex lens: $\frac{1}{f_{lens}} = (1.5 - 1)\frac{1}{R} = \frac{0.5}{R}$ The oil forms a bi-concave lens: $\frac{1}{f_{oil}} = (1.7 - 1)\left(-\frac{2}{R}\right) = -\frac{1.4}{R}$ $\frac{1}{f_{eq}} = 2 \times \frac{0.5}{R} - \frac{1.4}{R} = -\frac{0.4}{R}$ $f_{eq} = -\frac{R}{0.4} = -\frac{20}{0.4} = -50$ cm

Solution: For the lens: $u = -30$ cm, $f = 20$ cm $\frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{1}{60} \Rightarrow v = 60$ cm For the final image to coincide with the object, the rays must strike the convex mirror normally — i.e. the refracted rays must be directed towards the mirror's centre of curvature. Distance of mirror from lens $= 60 - R = 60 - 10 = 50$ cm

Solution: Object distance, $u = -1.5f$ By mirror formula, $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ $\frac{1}{v} + \frac{1}{-1.5f} = \frac{1}{-f}$ $\frac{1}{v} = -\frac{1}{f} + \frac{1}{1.5f} = -\frac{1}{f}\left(1 - \frac{2}{3}\right) = -\frac{1}{3f}$ $\Rightarrow v = -3f$

Solution: Key Concept: The net displacement of the image equals the difference between the image distances in the two cases. Case 1: $u_1 = -40$ cm, $f = -15$ cm $\frac{1}{f} = \frac{1}{v_1} + \frac{1}{u_1}$ gives $\frac{1}{v_1} = -\frac{1}{15} + \frac{1}{40} = \frac{-5}{120}$, so $v_1 = -24$ cm Case 2: $u_2 = -20$ cm $\frac{1}{v_2} = -\frac{1}{15} + \frac{1}{20} = \frac{-1}{60}$, so $v_2 = -60$ cm Displacement of image $= v_2 - v_1 = -60 - (-24) = -36$ cm $= 36$ cm, away from the mirror.

Solution: When light is incident normally on a plane mirror, it is reflected back along the same path. When the mirror is rotated by a small angle $\theta$, the reflected ray rotates by $2\theta$. The spot moves a distance y on a scale at distance x, so $2\theta = \frac{y}{x}$ $\Rightarrow \theta = \frac{y}{2x}$