NEET Practice Questions (MCQs) with Answers & Solutions

In Young's double-slit experiment, the position of the m-th maxima is given by:

\[ y_m = rac{m imes eta imes ext{distance between slits and screen}}{d} \]

where \( eta \) is the fringe width and d is the distance between the slits. For two different wavelengths \( \\lambda_1 \) and \( \\lambda_2 \), the ratio of the distances of the seventh maxima can be given by:

\[ rac{d_1}{d_2} = rac{eta_1}{eta_2} = rac{rac{ ext{distance between slits and screen} imes \\lambda_1}{d}}{rac{ ext{distance between slits and screen} imes \\lambda_2}{d}} = rac{\\lambda_1}{\\lambda_2} \]

Thus, the ratio \( \\frac{d_1}{d_2} \) is equal to \( \\frac{\\lambda_1}{\\lambda_2} \).

$ Use K-E = { hc \over \lambda } $

In Young's double slit experiment, the phase difference between the waves reaching the nth bright fringe is given by $n imes 2 ext{Ï€}$. For the 3rd bright fringe, $n = 3$, so the phase difference is $3 imes 2 ext{Ï€} = 6 ext{Ï€}$. Therefore, the correct answer is $6 ext{Ï€}$.

$ use \;n \lambda_1 = ( n+1) { \lambda \over 2 } $ $ \therefore n = 4 $

Diffraction is more pronounced for waves with larger wavelengths. Among the given options, radio waves have the longest wavelength. Therefore, they will undergo the maximum diffraction.

The angular width of the central maximum in a single-slit diffraction pattern is given by $2 heta = rac{2 ext{λ}}{a}$, where $ ext{λ}$ is the wavelength and $a$ is the slit width. Substituting $ ext{λ} = 6000 ext{Å}$ and $a = 12 imes 10^{-7} ext{cm}$, we get an angular width of approximately $60^ ext{°}$.

Polarization refers to the orientation of the oscillations of light waves. One of the key effects of polarization is the change in the intensity of light. When light is polarized, its intensity can vary depending on the angle and the method of polarization. Therefore, the correct answer is that intensity changes in the polarization phenomenon of light.

Fringe visibility (V) in an interference pattern is given by the formula: V = (I_max - I_min) / (I_max + I_min). For two coherent sources of equal intensity β, the fringe visibility is calculated using V = (2√β) / (1 + β). Hence, the correct option is $\frac{2\sqrt{\beta}}{1+\beta}$.