JEE MAIN - Physics (2021 - 31st August Evening Shift - No. 24)
In a Young's double slit experiment, the slits are separated by 0.3 mm and the screen is 1.5 m away from the plane of slits. Distance between fourth bright fringes on both sides of central bright is 2.4 cm. The frequency of light used is ______________ $$\times$$ 1014 Hz.
Answer
5
Explanation
8$$\beta$$ = 2.4 cm
$${{8\lambda \Delta } \over d}$$ = 2.4 cm
$${{8 \times 1.5 \times c} \over {0.3 \times {{10}^{ - 3}} \times f}} = 2.4 \times {10^{ - 2}}$$
f = 5 $$\times$$ 1014 Hz
$${{8\lambda \Delta } \over d}$$ = 2.4 cm
$${{8 \times 1.5 \times c} \over {0.3 \times {{10}^{ - 3}} \times f}} = 2.4 \times {10^{ - 2}}$$
f = 5 $$\times$$ 1014 Hz
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