Question 61:
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 62:
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 63:
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 64:
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 65:
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 66:
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 67:
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 68:
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 69:
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 70:
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 $n=3$, $x_3 = \frac{3 \times 5000 \times 10^{-10} \times 2}{0.2 \times 10^{-3}} = \frac{3 \times 10^{-6}}{0.2 \times 10^{-3}} = 15 \times 10^{-3} m = 1.5 cm$.