NEET Optical Instruments MCQs & PYQs

A rainbow is formed due to the dispersion of sunlight by water droplets, followed by total internal reflection and refraction. Therefore, total internal reflection and dispersion is the correct combined effect.

The blue colour of the sky is due to Rayleigh scattering. Since blue light has a shorter wavelength, it is scattered much more strongly by atmospheric particles than longer wavelengths like red.

To see objects at infinity, the correcting lens must form an image at the person's far point, so $u = -\infty$ and $v = -400 cm = -4 m$. Power $P = \frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{-4} - \frac{1}{-\infty}$. $P = -0.25 D$. A negative power indicates a concave lens.

The total converging power of the eye is $P = P_1 + P_2 = 40 D + 20 D = 60 D$. The focal length is $f = \frac{1}{P} = \frac{1}{60} m = \frac{100}{60} cm = \frac{5}{3} cm$. The distance between the retina and the cornea is equivalent to this focal length, which is $\approx 1.67 cm$.

The magnifying power of a microscope is inversely proportional to the focal length of its objective ($M \propto 1/f_o$). The magnifying power of a telescope is directly proportional to the focal length of its objective ($M = f_o/f_e$). Therefore, increasing $f_o$ decreases microscope magnification and increases telescope magnification.

Objective magnification is $m_o = \frac{f_o}{f_o + u_o}$. Substituting $f_o = 0.25$ and $u_o = -1/3.8$, $m_o = \frac{1/4}{1/4 - 1/3.8} = \frac{1/4}{(3.8 - 4)/15.2} = -19$. Total magnification is $M = m_o \times m_e$, so $95 = 19 \times m_e$ (using magnitudes). This gives the magnification of the eyepiece as $m_e = 5$.

A large aperture objective collects more light, which increases the light gathering power, making faint distant objects visible. It also inherently increases the resolving power of the telescope, allowing finer details to be resolved. Therefore, all the given statements are correct.

For the objective, $u_o = -200 cm$, $f_o = +40 cm$. Using $\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}$, we get $\frac{1}{v_o} = \frac{1}{40} - \frac{1}{200} = \frac{4}{200}$, so $v_o = 50 cm$. In normal adjustment, the intermediate image forms at the focus of the eyepiece, so $u_e = f_e = 4 cm$. The required separation is $L = v_o + f_e = 50 + 4 = 54 cm$.

Exposure time $t$ is directly proportional to the square of the f-number, so $t \propto (f-number)^2$. Thus, $\frac{t_2}{t_1} = \left(\frac{5.6}{2.8}\right)^2 = 2^2 = 4$. The new exposure time is $t_2 = 4 \times \frac{1}{200} = \frac{1}{50} = 0.02 second$.

Ray optics is treated as a limiting case of wave optics when the wavelength of light is negligible compared to the size of the objects or obstacles. It remains valid as long as the characteristic dimensions are much larger than the wavelength of light.