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PhysicsEasyMCQ2021 · 24 Feb Shift 1

Q17.In a Young's double slit experiment, the width of the one of the slit is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit-width. Find the ratio of the maximum to the minimum intensity in the interference pattern. JEE Main 2021 (24 Feb Shift 1) JEE Main Previous Year Paper (1) 1: 4 (2) 2: 1 (3) 3: 1 (4) 4: 1

What This Question Tests

This is a direct recall question about the fundamental relationship between the focal length and the radius of curvature for a spherical mirror.

Concepts Tested

Focal length of spherical mirrorRadius of curvature

Formulas Used

f = R/2

📚 NCERT Sections This Tests

10.4In A Young’S Double-Slit Experiment, The Slits Are Separated By

Physics Class 12 · Chapter 10

77% match

10.4 In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.

10.5In Young’S Double-Slit Experiment Using Monochromatic Light Of

Physics Class 12 · Chapter 10

77% match

10.5 In Young’s double-slit experiment using monochromatic light of wavelength l, the intensity of light at a point on the screen where path difference is l, is K units. What is the intensity of light at a point where path difference is l/3?

9.17(A) Sin I¢C = 1.44/1.68 Which Gives I¢C = 59°. Total Internal Reflection

Physics Class 12 · Chapter 9

73% match

9.17 (a) sin i¢c = 1.44/1.68 which gives i¢c = 59°. Total internal reflection takes place when i > 59° or when r < rmax = 31°. Now, (sin i /sin r max max ) = 1.68 , which gives imax ~ 60°. Thus, all incident rays of angles in the range 0 < i < 60° will suffer total internal reflections in the pipe. (If the length of the pipe is finite, which it is in practice, there will be a lower limit on i determined by the ratio of the diameter to the length of the pipe.) (b) If there is no outer coating, i¢c = sin–1(1/1.68) = 36.5°. Now, i = 90° will have r = 36.5° and i¢ = 53.5° which is greater than i¢c. Thus, all incident rays (in the range 53.5° < i < 90°) will suffer total internal reflections.