1
MHT CET 2021 20th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

In Young's double slit experiment, the '$$\mathrm{n^{th}}$$' maximum of wavelength '$$\lambda_1$$' is at a distance '$$\mathrm{y_1}$$' from the central maximum. When the wavelength of the source is changed to '$$\lambda_2$$', $$\left(\frac{\mathrm{n}}{2}\right)^{\text {th }}$$ maximum is at a distance of '$$\mathrm{y_2}$$' from its central maximum. The ratio $$\frac{y_1}{y_2}$$ is

A
$$\frac{\lambda_2}{2 \lambda_1}$$
B
$$\frac{2 \lambda_1}{\lambda_2}$$
C
$$\frac{2 \lambda_2}{\lambda_1}$$
D
$$\frac{\lambda_1}{2 \lambda_2}$$
2
MHT CET 2021 20th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

Light of wavelength '$$\lambda$$' is incident on a single slit of width 'a' and the distance between slit and screen is 'D'. In diffraction pattern, if slit width is equal to the width of the central maximum then $$\mathrm{D}=$$

A
$$\frac{a^2}{\lambda}$$
B
$$\frac{\mathrm{a}}{\lambda}$$
C
$$\frac{a^2}{2 \lambda}$$
D
$$\frac{\mathrm{a}}{2 \lambda}$$
3
MHT CET 2021 20th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

In Fraunhofer diffraction pattern, slit width is 0.2 mm and screen is at 2m away from the lens. If wavelength of light used is 5000$$\mathop A\limits^o $$ then the distance between the first minimum on either side of the central maximum is ($$\theta$$ is small and measured in radian)

A
2 $$\times$$ 10$$^{-2}$$ m
B
10$$^{-1}$$ m
C
10$$^{-2}$$ m
D
10$$^{-3}$$ m
4
MHT CET 2020 16th October Morning Shift
MCQ (Single Correct Answer)
+1
-0

In diffraction experiment, from a single slit, the angular width of the central maxima does not depend upon

A
ratio of wavelength and slit width
B
distance of the slit from the screen
C
wavelength of light used
D
width of the slit
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