NEET Ray Optics MCQs & PYQs

Assuming refractive index $n=1.5$, focal length $\frac{1}{f} = (1.5-1)(\frac{1}{20} - \frac{1}{-20}) = \frac{1}{20} \Rightarrow f = 20 cm$. Using lens formula $\frac{1}{v} - \frac{1}{-30} = \frac{1}{20}$, we get $v = +60 cm$. Magnification $m = \frac{v}{u} = \frac{60}{-30} = -2$. Image height $h_i = m h_o = -2 \times 2 = -4 cm$. The image is real, inverted, and 4 cm high.

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.

When light travels from a rarer medium like vacuum to a denser medium, its frequency remains unchanged while its velocity decreases. Since velocity $v = f \lambda$, the wavelength also decreases. Specifically, $\lambda' = \frac{\lambda}{\mu}$, making the new wavelength smaller.

The wavelength of light in a medium is given by $\lambda' = \frac{\lambda}{\mu}$. Substituting the given values, the wavelength in glass is $\lambda' = \frac{5460 \AA}{1.5} = 3640 \AA$.

For a star approaching the earth at a speed much less than $c$, the change in wavelength (Doppler shift) is $\Delta\lambda = \lambda \frac{v}{c}$. Substituting the values, $\Delta\lambda = 5000 \times \frac{1.5 \times 10^4}{3 \times 10^8} = 5000 \times 0.5 \times 10^{-4} = 25 \AA$.