Question

Q] In a biprism experiment, the fringes are observed in the focal plane of the eyepiece at a distance of 1.2 m from the slits. The distance between the central bright band and the 20th bright band is 0.4 cm. When a convex lens is placed between the biprism and the eyepiece, 90 cm from the eyepiece, the distance between the two virtual magnified images is found to be 0.9 cm. Determine the wavelength of light used.​

Answer 1

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★GIVEN:-

• distance of 1.2 m from the slits.

• The distance between the central bright band and the 20th bright band is 0.4 cm.

• the distance between the two virtual magnified images is 0.9cm.

★TO FIND:-

• wavelength of light used.

★SOLUTION:-

$\therefore\displaystyle \sf {Y_{20} = 0.4cm = 0.4 \times {10}^{ - 2} m}$

$: \implies \displaystyle \sf{W = \frac{Y_{20}}{20} }$

$: \implies \displaystyle \sf{ \frac{0.4}{20} \times {10}^{ - 2} m }$

$: \implies \displaystyle \sf{2 \times {10}^{ - 4} m}$

$\leadsto \sf{d1 = 0.9 cm }$

$\leadsto \sf{0.9 \times {10}^{ - 2}m }$

$\leadsto \sf{90cm = 0.9m}$

$\therefore \displaystyle \sf{u_1 = D - v_1 = 1.2m - 0.9m = 0.3m}$

Now,

$: \implies \displaystyle \sf{ \frac{d_1}{d} = \frac{v_1}{u_1} }$

$\therefore \displaystyle \sf{d = \frac{d_1 \: u_1}{v_1} }$

$\therefore \displaystyle \sf{d = \frac{(0.9 \times {10}^{ - 2})(0.3) }{0.9} m }$

$\longrightarrow\displaystyle \sf{d = 3 \times {10}^{ - 3} m}$

The wavelength of light,

$\displaystyle \sf{ λ = ( \frac{Wd}{D} )}$

$\longrightarrow \displaystyle \sf{ λ = \frac{2 \times {10}^{ - 4} \times 3 \times {10}^{ - 3} }{1.2}m }$

$\longrightarrow \displaystyle \sf{5 \times {10}^{ - 7}m }$

$\longrightarrow \displaystyle \sf{5 \times {10}^{ - 7} \times {10}^{10} Å }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \displaystyle { \boxed{ \underline{ \huge{ \color{magenta}{\sf{ \bf{5 000Å }}}}}}}$

Answer 2

Explanation:

i attached one picture please refer it