Question 91:
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 92:
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 93:
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 94:
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 95:
easyTwo slits in Young’s experiment have widths in the ratio $1:25$. The ratio of intensity at the maxima and minima in the interference pattern, $\frac{I_{max}}{I_{min}}$ is:
(2015 Re)
Ratio of slit widths $W_1/W_2 = I_1/I_2 = 1/25$. Amplitudes ratio $a_1/a_2 = \sqrt{1/25} = 1/5$. The intensity ratio is $\frac{I_{max}}{I_{min}} = \left(\frac{a_1+a_2}{a_1-a_2}\right)^2 = \left(\frac{1+5}{1-5}\right)^2 = \left(\frac{6}{-4}\right)^2 = \frac{36}{16} = \frac{9}{4}$.
Question 96:
easyIn the Young’s double-slit experiment, the intensity of light at a point on the screen where the path difference is $\lambda$ is K, ($\lambda$ being the wavelength of light used). The intensity at a point where the path difference is $\lambda/4$, will be:
(2014)
Phase difference $\phi = \frac{2\pi}{\lambda} \times \Delta x$. When $\Delta x = \lambda$, $\phi = 2\pi$, intensity $I = I_{max}\cos^2(\pi) = I_{max} = K$. When $\Delta x = \lambda/4$, $\phi = \pi/2$. The intensity is $I = I_{max}\cos^2(\phi/2) = K\cos^2(\pi/4) = K(1/\sqrt{2})^2 = K/2$.
Question 97:
easyIn Young’s double slit experiment, the slits are 2 mm apart and are illuminated by photons of two wavelengths $\lambda_1 = 12000 \AA$ and $\lambda_2 = 10000 \AA$. At what minimum distance from the common central bright fringe on the screen 2 m from the slit will a bright fringe from one interference pattern coincide with a bright fringe from the other?
(2013)
For coincidence, $n_1\lambda_1 = n_2\lambda_2 \Rightarrow n_1(12000) = n_2(10000) \Rightarrow \frac{n_1}{n_2} = \frac{5}{6}$. Minimum values are $n_1=5$, $n_2=6$. The minimum distance $y = \frac{n_1\lambda_1 D}{d} = \frac{5 \times 12000 \times 10^{-10} \times 2}{2 \times 10^{-3}} = 6 \times 10^{-3} m = 6 mm$.
Question 98:
easyInterference was observed in interference chamber where air was present, now the chamber is evacuated, and if the same light is used, a careful observer will see
(1993)
When the chamber is evacuated, the refractive index decreases (from $\mu_{air}$ to 1). The wavelength of light $\lambda = \lambda_0/\mu$ increases. Since fringe width $\beta = \frac{\lambda D}{d}$, an increase in wavelength leads to a larger fringe width.
Question 99:
easyIf yellow light emitted by sodium lamp in Young’s double slit experiment is replaced by monochromatic blue light of the same intensity
(1992)
Fringe width is given by $\beta = \frac{\lambda D}{d}$. Since the wavelength of blue light is less than that of yellow light ($\lambda_{blue} < \lambda_{yellow}$), the fringe width will decrease.
Question 100:
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 $\lambda = 0.6 \times 10^{-6} m = 6 \times 10^{-5} cm$.