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Wave Optics

JEE Main level, numerical answer, medium

88 of 191

Question 88 of 191

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

Show answer and solution

Answer: 600

\beta =\dfrac{\lambda D}{d}

\lambda =\dfrac{\beta d}{D}

\lambda =\dfrac{6\times {10}^{-3}\times {10}^{-3}}{10}

\lambda =6\times {10}^{-7}m=600\times {10}^{-9}m

\lambda =600nm

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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More Physics practice

JEE Main Physics practice question on Wave Optics (Youngs Double Slit Experiment), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.

A fringe width of 6mm6mm6mm was produced for two slits separated by 1mm1mm1mm apart. The screen is placed 10m10m10m away. The wavelength of light used is xxx nmnmnm. The value of xxx to the nearest integer is ______.

JEE Main question paper PDF

More Physics practice

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