In Young's double slit experiment, 8^{\text{th }} maximum with wavelength ' \lambda_1 ' is at a distance ' d_1 ' from the central maximum and 6^{\text{th }} maximum with wavelength ' \lambda_2 ' is at a distance ' {\d}_2 '. Then { {\d}_2}/{ {\d}_1} is

\frac{3 \lambda_1}{4 \lambda_2}

\frac{3 \lambda_2}{4 \lambda_1}

\frac{4 \lambda_1}{3 \lambda_2}

\frac{4 \lambda_2}{3 \lambda_1}

MHTCET 14 May 2023 Shift 1 Solved Paper

Section: Physics

Question : 138 of 150

Marks: +1, -0

In Young's double slit experiment, 8‌th ‌ maximum with wavelength ' λ1 ' is at a distance ' d1 ' from the central maximum and 6‌th ‌ maximum with wavelength ' λ2 ' is at a distance ' d2 '. Then ‌

‌
3λ1
4λ2
‌
3λ2
4λ1
‌
4λ1
3λ2
‌
4λ2
3λ1

Validate

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