Question 121:
easyRatio of intensities of two waves are given by 4 : 1. Then ratio of the amplitudes of the two waves is:
(1991)
The intensity of a wave is directly proportional to the square of its amplitude ($I \propto A^2$). Therefore, the ratio of amplitudes is $A_1 / A_2 = \sqrt{I_1 / I_2} = \sqrt{4 / 1} = 2 / 1$.
Question 122:
easyIn a Young’s double slit experiment, a student observes 8 fringes in a certain segment of screen when a monochromatic light of 600 nm wavelength is used. If the wavelength of light is changed to 400 nm, then the number of fringes he would observe in the same region of the screen is
(2022)
The width of the screen segment remains constant. Thus, $n_1 \beta_1 = n_2 \beta_2$, which implies $n_1 \lambda_1 = n_2 \lambda_2$. Substituting the values: $8 \times 600 = n_2 \times 400$, giving $n_2 = 12$ fringes.
Question 123:
easyIn Young’s double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes:
(2020)
Fringe width is given by $beta = frac{lambda D}{d}$. When $D' = 2D$ and $d' = d/2$, the new fringe width is $beta' = frac{lambda (2D)}{(d/2)} = 4 frac{lambda D}{d} = 4beta$. It becomes four times the original.
Question 124:
easyTwo coherent sources of light interfere and produce fringe pattern on a screen. For central maximum, the phase difference between the two waves will be.
(2020-Covid)
At the central maximum, the path difference between the two interfering waves is zero. Consequently, the phase difference $\Delta\phi = \frac{2\pi}{\lambda} \Delta x$ is also zero.
Question 125:
easyIn a double slit experiment, when light of wavelength 400 nm was used, the angular width of the first minima formed on a screen placed 1 m away, was found to be $0.2^\circ$. What will be the angular width of the first minima, if the entire experimental apparatus is immersed in water? ($\mu_{water} = 4/3$)
(2019)
The angular width is $\theta = \frac{\lambda}{d}$. In a medium of refractive index $\mu$, the wavelength becomes $\lambda' = \frac{\lambda}{\mu}$. Therefore, the new angular width is $\theta' = \frac{\theta}{\mu} = \frac{0.2^\circ}{4/3} = 0.15^\circ$.
Question 126:
easyIn Young’s double slit experiment the separation d between the slits is 2 mm, the wavelength $\lambda$ of the light used is $5896 AA$ and distance D between the screen and slits is 100 cm. It is found that the angular width of the fringes is $0.20^\circ$. To increase the fringe angular width to $0.21^\circ$ (with same $\lambda$ and D) the separation between the slits needs to be changed to
(2018)
The angular fringe width is $\theta = \frac{\lambda}{d}$, which means $\theta \propto \frac{1}{d}$. So, $\theta_1 d_1 = \theta_2 d_2$. Substituting the values: $0.20 \times 2 = 0.21 \times d_2$. This yields $d_2 = \frac{0.40}{0.21} \approx 1.9 mm$.
Question 127:
moderateYoung’s double slit experiment is first performed in air and then in a medium other than air. It is found that $8^{th}$ bright fringe in the medium lies where $5^{th}$ dark fringe lies in air. The refractive index of the medium is nearly:
(2017-Delhi)
Position of the 8th bright fringe in medium is $y_8 = 8 \frac{\lambda_{med} D}{d}$. Position of the 5th dark fringe in air is $y_5 = (5 - 0.5) \frac{\lambda_{air} D}{d} = 4.5 \frac{\lambda_{air} D}{d}$. Equating them: $8 \lambda_{med} = 4.5 \lambda_{air}$. Since $\lambda_{med} = \frac{\lambda_{air}}{\mu}$, we get $\frac{8}{\mu} = 4.5$, so $\mu = \frac{8}{4.5} \approx 1.78$.
Question 128:
difficultThe intensity at the maximum in a Young’s double slit experiment is $I_0$. Distance between two slits is $d = 5\lambda$, where $lambda$ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance $D = 10d$?
(2016-I)
The point in front of one slit has $y = d/2$. The path difference is $Delta x = frac{yd}{D} = frac{(d/2)d}{10d} = frac{d}{20}$. Since $d = 5lambda$, $Delta x = frac{5lambda}{20} = frac{lambda}{4}$. The phase difference is $Deltaphi = frac{2pi}{lambda} Delta x = frac{pi}{2}$. The intensity is $I = I_0 cos^2(frac{Deltaphi}{2}) = I_0 cos^2(frac{pi}{4}) = frac{I_0}{2}$.
Question 129:
moderateIn Young’s double slit experiment carried out with light of wavelength ($\lambda$) = $5000 \AA$, the distance between the slits is 0.2 mm and the screen is at 200 cm from the slits. The central maximum is at x = 0. The third maximum (taking the central maximum as zeroth maximum) will be at x equal to
(1992)
Position of nth maximum is $x_n = \frac{n\lambda D}{d}$. For $n=3$, $x_3 = \frac{3 \times 5000 \times 10^{-10} \times 2}{0.2 \times 10^{-3}} = \frac{3 \times 10^{-6}}{0.2 \times 10^{-3}} = 15 \times 10^{-3} m = 1.5 cm$.
Question 130:
easyIn Young’s experiment, two coherent sources are placed 0.90 mm apart and fringe are observed one metre away. If it produces second dark fringe at a distance of 1 mm from central fringe, the wavelength of monochromatic light is used would be:
(1991)
Position of nth dark fringe is $y_n = \frac{(2n-1)\lambda D}{2d}$. For 2nd dark fringe ($n=2$), $y = \frac{3\lambda D}{2d}$. Given $y = 1 mm = 10^{-3} m$, so $10^{-3} = \frac{3 \times \lambda \times 1}{2 \times 0.90 \times 10^{-3}}$. Solving gives $\lambda = 0.6 \times 10^{-6} m = 6 \times 10^{-5} cm$.