Optical Instruments — NEET Physics PYQ MCQs with Solutions

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: For the objective: $u_0 = -200$ cm, $f_0 = 40$ cm $\frac{1}{v} = \frac{1}{40} - \frac{1}{200} = \frac{4}{200} \Rightarrow v = 50$ cm The image forms at the first focus of the eyepiece, so the separation is $v + f_e = 50 + 4 = 54$ cm

Solution: The image of an object at infinity must form at the person's far point (400 cm = 4 m): $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ with $u = \infty$, $v = -4$ m $f = -4$ m $P = \frac{1}{f} = -0.25$ D — a concave lens of power −0.25 diopter.

Solution: The objective (with the line of length L) acts as an object for the eyepiece placed at distance $f_o + f_e$. The image length I satisfies $\frac{I}{L} = \frac{f_e}{f_o}$ Magnification of telescope $M = \frac{f_o}{f_e} = \frac{L}{I}$

Solution: For a microscope: $m = \frac{L}{f_o}\cdot\frac{D}{f_e}$, so $m \propto \frac{1}{f_o}$ — magnifying power decreases. For a telescope: $m = \frac{f_o}{f_e}$, so $m \propto f_o$ — magnifying power increases.

Solution: Total power $P_{eq} = P_1 + P_2 = 40 + 20 = 60$ D For a distant object the image forms at the focus, i.e. on the retina: $f_{eq} = \frac{100}{P_{eq}} = \frac{100}{60} = 1.67$ cm

Solution: Magnification: $\frac{f_o}{f_e} = 9$, so $f_o = 9f_e$ Length: $f_o + f_e = 20$ $9f_e + f_e = 20 \Rightarrow f_e = 2$ cm, $f_o = 18$ cm

Solution: Apparent depth of the mark through the slab $= \frac{3}{1.5} = 2$ cm The mark appears raised by $3 - 2 = 1$ cm, so the microscope must be moved 1 cm upward.

Solution: Resolving limit: $\frac{x}{D} = \frac{\lambda}{d}$ $x = \frac{\lambda D}{d} = \frac{5000 \times 10^{-10} \times 1000}{0.1} = 5 \times 10^{-3}$ m $= 5$ mm

Solution: Angular limit of resolution: $\theta = \frac{\lambda}{d} = \frac{y}{D}$ $y = \frac{\lambda D}{d} = \frac{5 \times 10^{-7} \times 50}{2 \times 10^{-3}} = 12.5 \times 10^{-3}$ m $= 1.25$ cm

Solution: $m = \frac{f_o}{f_e} = 10$ and $L = f_o + f_e = 44$ cm $f_o + \frac{f_o}{10} = 44 \Rightarrow \frac{11f_o}{10} = 44$ $f_o = 40$ cm

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.