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QJEEM21Ph5ea9: JEE Main 2021 Physics - Class 12 JEEM21Ph5ea9 Solution: A fringe width of 6 mm was produced for two slits separated by…

A fringe width of 6 mm was produced for two slits separated by 1 mm apart. The screen is placed 10 m away. The wavelength of light used is 'x' nm. The value of 'x' to the nearest integer is ____________.

Step-by-Step Verified Solution

β\beta = 6 mm, d = 1 mm, D = 10 m

λ\lambda = ?

We know, β=λDd\beta = {{\lambda D} \over d}

6 ×\times 10 −- 3 = λ×101×10−3{{\lambda \times 10} \over {1 \times {{10}^{ - 3}}}}

∴\therefore λ\lambda = 6×10−3×1×10−310{{6 \times {{10}^{ - 3}} \times 1 \times {{10}^{ - 3}}} \over {10}}

λ\lambda = 600 ×\times 10 −- 9 m

∴\therefore λ\lambda = 600 nm

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