Question 1:
moderateWhich one of the following phenomena is not explained by Huygen’s construction of wavefront?
(1988)
Huygen's wave theory of light successfully explained phenomena like reflection, refraction, interference, and diffraction. However, it could not explain the origin of spectra, which requires quantum mechanics and the particle nature of light.
Question 2:
moderateThe interference pattern is obtained with two coherent light sources of intensity ratio n. In the interference pattern, the ratio $\frac{I_{max} – I_{min}}{I_{max} + I_{min}}$ will be:
(2016 – II)
The maximum and minimum intensities are $I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2$ and $I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2$. Given $I_1/I_2 = n$, we can write $I_1 = n I_2$. The required ratio is $\frac{(\sqrt{n}+1)^2 - (\sqrt{n}-1)^2}{(\sqrt{n}+1)^2 + (\sqrt{n}-1)^2}$. Expanding the squares gives $\frac{4\sqrt{n}}{2(n+1)} = \frac{2\sqrt{n}}{n+1}$.
Question 3:
moderateTwo periodic waves of intensities $I_1$ and $I_2$ pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is:
(2008)
The maximum intensity is $I_{max} = I_1 + I_2 + 2\sqrt{I_1 I_2}$ and the minimum intensity is $I_{min} = I_1 + I_2 - 2\sqrt{I_1 I_2}$. Their sum is $I_{max} + I_{min} = 2(I_1 + I_2)$.
Question 4:
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 5:
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 6:
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 7:
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 8:
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 9:
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 10:
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 $\mu = \frac{8}{4.5} \approx 1.78$.