Lenses — NEET Physics PYQ MCQs with Solutions

Solution: For the emerging beam to remain parallel, the net focal length of the combination must be infinite: $\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}$ $\frac{1}{\infty} = \frac{1}{20} - \frac{1}{5} + \frac{d}{20 \times 5}$ $\frac{3}{20} = \frac{d}{100} \Rightarrow d = 15$ cm (Equivalently, the focus of A coincides with the focus of B: $d = 20 - 5 = 15$ cm.)

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: 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: $f = \frac{1}{P} = \frac{1}{10}$ m $= 10$ cm, with $R_1 = 10$ cm, $R_2 = -10$ cm By the lens maker's formula: $\frac{1}{10} = (\mu - 1)\left(\frac{1}{10} + \frac{1}{10}\right) = (\mu - 1)\frac{2}{10}$ $\mu - 1 = \frac{1}{2} \Rightarrow \mu = \frac{3}{2}$

Solution: When a lens is cut along a plane containing the principal axis, the radii of curvature of both surfaces remain unchanged, so the focal length — and hence the power — of each half remains the same as the original lens: $P = \frac{1}{f}$.

Solution: Let $R_1 = R$ and $R_2 = -2R$. Using the lens maker's formula: $\frac{1}{25} = (1.5 - 1)\left(\frac{1}{R} + \frac{1}{2R}\right) = 0.5 \times \frac{3}{2R}$ $R = \frac{3}{2} \times 25 \times 0.5 = 18.75$ cm $R_1 = 18.75$ cm, $R_2 = 2 \times 18.75 = 37.5$ cm

Solution: Case I — lenses in contact: $\frac{1}{F_1} = \frac{1}{f} + \frac{1}{f} = \frac{2}{f} \Rightarrow F_1 = \frac{f}{2}$ Case II — glycerine (same $\mu$ as glass) between the lenses forms a bi-concave liquid lens of focal length $-f$: $\frac{1}{F_2} = \frac{1}{f} - \frac{1}{f} + \frac{1}{f} = \frac{1}{f} \Rightarrow F_2 = f$ $F_1 : F_2 = 1 : 2$

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: $\frac{1}{f_1} = (\mu_1 - 1)\left(\frac{1}{\infty} - \frac{1}{-R}\right) = \frac{\mu_1 - 1}{R}$ $\frac{1}{f_2} = (\mu_2 - 1)\left(\frac{1}{-R} - \frac{1}{\infty}\right) = -\frac{\mu_2 - 1}{R}$ $\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{\mu_1 - \mu_2}{R}$ $f = \frac{R}{\mu_1 - \mu_2}$

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: Taking $\mu = 1.5$: $\frac{1}{f} = (1.5 - 1)\times\frac{2}{20} = \frac{1}{20}$, so $f = 20$ cm With $u = -30$ cm: $\frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{1}{60} \Rightarrow v = 60$ cm $m = \frac{v}{u} = \frac{60}{-30} = -2$ $h_i = -2 \times 2 = -4$ cm The image is real, inverted and 4 cm high.

Solution: Intensity is proportional to the exposed area of the aperture: $\frac{I_2}{I_1} = \frac{\frac{\pi d^2}{4} - \frac{\pi (d/2)^2}{4}}{\frac{\pi d^2}{4}} = \frac{3}{4}$ So the intensity becomes $\frac{3I}{4}$, while the focal length remains unchanged (f).

Solution: The image of the sun forms at the focus, so $v = 0.1$ m. $\frac{I}{O} = \frac{v}{u}$ $I = O \times \frac{v}{u} = 1.39 \times 10^9 \times \frac{0.1}{1.5 \times 10^{11}} = 9.2 \times 10^{-4}$ m

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: From the lens maker's formula in a liquid: $\frac{1}{f_l} = \left(\frac{\mu_g}{\mu_l} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$ With $\mu_g = \mu_l$ the bracket $\left(\frac{\mu_g}{\mu_l} - 1\right) = 0$, so $\frac{1}{f_l} = 0$ and the focal length becomes infinite.

Solution: Cut along the principal axis (XOX′): each half keeps both radii of curvature $R$ and $-R$, so $f' = f$. Cut perpendicular to the principal axis (YOY′): each half becomes plano-convex with $R_1 = R$, $R_2 = \infty$: $\frac{1}{f''} = (\mu - 1)\frac{1}{R} = \frac{1}{2f}$ Hence $f' = f$, $f'' = 2f$.

Solution: For an image of equal size, the object must be at the centre of curvature (2f) of the lens, i.e. $|u| = |v| = d$. $\frac{1}{f} = \frac{1}{d} + \frac{1}{d} = \frac{2}{d}$ $f = \frac{d}{2}$

Solution: The silvered lens acts as a mirror of focal length F, where $\frac{1}{F} = \frac{2}{f_l} + \frac{1}{f_m}$ For the plane mirror $f_m = \infty$; for the lens $\frac{1}{f_l} = \frac{\mu - 1}{R}$ $\frac{1}{F} = \frac{2(\mu - 1)}{R} \Rightarrow F = \frac{R}{2(\mu - 1)} = \frac{20}{2 \times 0.5} = 20$ cm

Solution: For a planoconvex lens, $R_2 = \infty$: $\frac{1}{f} = (\mu - 1)\frac{1}{R_1} = (1.6 - 1)\times\frac{1}{60} = \frac{0.6}{60}$ $f = \frac{60}{0.6} = 100$ 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: From the lens maker's formula $\frac{1}{f} \propto (\mu - 1)$, and by Cauchy's formula $\mu \propto \frac{1}{\lambda}$, so $f \propto \lambda$. Focal length is maximum for red (longest wavelength) and minimum for violet, i.e. $f_V < f_R$.

Solution: With focal lengths in cm, $P = \frac{100}{f}$ D: $P = \frac{100}{80} - \frac{100}{50} = 1.25 - 2 = -0.75$ D

Solution: In the displacement method the magnifications in the two positions satisfy $m_1 m_2 = 1$ for linear size, so the area of the source is the geometric mean of the two image areas: $A = \sqrt{A_1 A_2}$

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: $\frac{f_l}{f_a} = \frac{\mu_g - 1}{\frac{\mu_g}{\mu_l} - 1} = \frac{1.5 - 1}{\frac{1.5}{1.25} - 1} = \frac{0.5}{0.2} = \frac{5}{2}$ $f_l = \frac{5}{2} \times 2 = 5$ cm