Wave Optics MCQs for NEET — Physics Questions with Answers

$ For, plano - convex lens , { 1 \over f_1 } = { 1 \over f_3 } = (n-1) \left( { 1 \over \infty } \times {1 \over R } \right) = {1 \over 24 }$ $ For , double convex lens \therefore { 1 \over f_1} + {1 \over f_2} +{ 1 \over f_3 }= { -1 \over 60 } $ $ \therefore {1 \over f_2} = { -1 \over 10 } $ $ now \therefore { 1 \over f_2 } = (n-1) \left( {1 \over R_1} - {1 \over R_2 }\right) = n =1.6 $

Huygens' wave theory of light can explain wave phenomena such as diffraction and interference. However, it cannot explain the photoelectric effect, which requires the concept of light as particles (photons) as explained by Einstein's quantum theory.

The fringe width (β) in a double-slit experiment is directly proportional to the wavelength (λ) of the light used. Given λ_r = 8000 Å and λ_ν = 4000 Å, the ratio of their fringe widths is β_r / β_ν = λ_r / λ_ν = 8000 Å / 4000 Å = 2:1.

When two coherent light waves of the same intensity (I) interfere constructively, the maximum intensity (I_max) is given by:

\[ I_{ ext{max}} = 4I \]

This is because the electric field amplitudes add up in constructive interference, resulting in the intensity being proportional to the square of the amplitude. Since the minimum intensity given is zero, the maximum intensity will be 4I.

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 \( \\frac{\\lambda_1}{\\lambda_2} \).

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 $60^ ext{°}$.