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QJEEM21Ph6031: JEE Main 2021 Physics - Class 12 JEEM21Ph6031 Solution: In a Young's double slit experiment, the slits are separated by …

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.

Step-by-Step Verified Solution

8 β\beta = 2.4 cm

8λΔd{{8\lambda \Delta } \over d} = 2.4 cm

8×1.5×c0.3×10−3×f=2.4×10−2{{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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