NEET Ray Optics MCQs & PYQs

Focal length of a lens is independent of its aperture, so it remains $f$ . Intensity is proportional to the exposed area. Original area $A = \frac{\pi d^2}{4}$ . Covered area $a = \frac{\pi (d/2)^2}{4} = \frac{A}{4}$ . Remaining area $A' = A - \frac{A}{4} = \frac{3A}{4}$ . New intensity is proportional to remaining area, so $I' = \frac{3I}{4}$ .

Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ . For a concave mirror, object distance $u = -1.5f$ and focal length is $-f$ . Substituting these values gives $\frac{1}{v} + \frac{1}{-1.5f} = \frac{1}{-f}$ , which implies $\frac{1}{v} = \frac{1}{1.5f} - \frac{1}{f} = \frac{1 - 1.5}{1.5f} = \frac{-0.5}{1.5f} = \frac{-1}{3f}$ . Thus, $v = -3f$ .

Initially $u_1 = -40 cm$ , $f = -15 cm$ . Using mirror formula, $v_1 = \frac{u_1 f}{u_1 - f} = \frac{(-40)(-15)}{-40 - (-15)} = \frac{600}{-25} = -24 cm$ . After moving 20 cm towards the mirror, $u_2 = -20 cm$ . The new image position is $v_2 = \frac{(-20)(-15)}{-20 - (-15)} = \frac{300}{-5} = -60 cm$ . Displacement is $v_2 - v_1 = -60 - (-24) = -36 cm$ , meaning 36 cm away from the mirror.

Negative magnification implies a real inverted image, which is only formed by a concave mirror. So A ( $m = -2$ ) and B ( $m = -1/2$ ) match with (2) Concave and (3) Real. Positive magnification > 1 ( $m = +2$ ) implies a virtual magnified image, formed only by a concave mirror (2, 4). Positive magnification < 1 implies a virtual diminished image, formed by a convex mirror (1, 4).

The rod's closer end is at $u_1 = -20 cm$ (which is $2f$ since $f = -10 cm$ ). The image of this end forms at the center of curvature, $v_1 = -20 cm$ . The farther end is at $u_2 = -30 cm$ . Its image is at $v_2 = \frac{u_2 f}{u_2 - f} = \frac{(-30)(-10)}{-30 - (-10)} = \frac{300}{-20} = -15 cm$ . The length of the image is $|v_1 - v_2| = |-20 - (-15)| = 5 cm$ .

Light from infinity passes through the convex lens and converges at its principal focus, a distance $f_2$ from the lens. For the light to return to infinity, it must retrace its path after reflection from the concave mirror. This happens if the rays strike the mirror normally, meaning they appear to come from the mirror's center of curvature (at distance $2f_1$ ). So, $d = 2f_1 + f_2$ .

For a person to see their full, complete image in a vertical plane mirror, the minimum length of the mirror required is exactly half of the person's height, regardless of their distance from the mirror. Therefore, minimum length = $6 / 2 = 3 feet$ .

Velocity of light in the material is given by $v = f \lambda = (2 \times 10^{14} Hz) \times (5000 \times 10^{-10} m) = 10^8 m/s$ . The refractive index of the material is calculated as $n = \frac{c}{v} = \frac{3 \times 10^8}{10^8} = 3.00$ .

When a light ray travels from one medium to another, its frequency remains absolutely unchanged because frequency is a fundamental characteristic of the source. Velocity and wavelength change based on the refractive index. Thus, $n = n'$ .

The speed of light in a given medium is related to the refractive index by $v = \frac{c}{n}$ . Given the speed of light in vacuum $c = 3 \times 10^8 m/s$ and the refractive index of water $n = 1.33 = \frac{4}{3}$ . We find $v = \frac{3 \times 10^8}{4/3} = \frac{9 \times 10^8}{4} = 2.25 \times 10^8 m/s$ .