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In a Young’s double-slit experiment using identical slits, when only one slit is used, the total energy reaching the screen is E0 and the intensity of light at any point on the screen is I0. When both slits are used, and fringes are formed on the screen, the total energy reaching the screen is E and the maximum intensity on the screen is I. Then, (a) E = 2E0, I = 2I0 (b) E = 4E0, I = 4I0 (c) E = 2E0, I = 4I0 (d) E = 4E0, I = 2I0 |
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Answer» Correct Answer is: (c) E = 2E0, I = 4I0 |
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