Question 11:
moderateIn the Young’s double slit experiment, the ratio of intensities of bright and dark fringes is 9. This means that :-
(A) The intensities of individual sources are 5 and 4 units respectively
(B) The intensities of individual sources are 4 and 1 units respectively
(C) The ratio of their amplitudes is 3
(D) The ratio of their amplitudes is 2
Question 12:
moderateIn a Young’s double slit experiment, the fringe width is found to be 0.4 mm. If the whole apparatus is immersed in water of refractive index 4/3 without disturbing the geometrical arrangement, the new fringe width will be :
Question 13:
moderateIn a Young’s double slit experiment the intensity at a point where the path difference is λ/6 (λ being the wavelength of the light used) is I. If I0 denotes the maximum intensity, I/I0 is equal to :
Question 14:
moderateIn the Young’s double slit experiment, for which colour the fringe width is least :
Question 15:
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 16:
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 17:
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 18:
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 19:
moderateThe Young’s double slit experiment is performed with blue and with green light of wavelengths $4360 \AA$ and $5460 \AA$ respectively. If x is the distance of 4th maxima from the central one, then:
(1990)
The distance of the nth maximum from the central maximum is $x = \frac{n\lambda D}{d}$. Thus, $x \propto \lambda$. Since $\lambda_{blue} < \lambda_{green}$, it follows that $x(blue) < x(green)$.
Question 20:
moderateInterference is possible in:
(1989)
Interference is a fundamental wave phenomenon based on the principle of superposition. It is exhibited by all types of waves, including both electromagnetic (light) waves and mechanical (sound) waves.