1.

Here We Go!!! Bichromatic Light Slit width is 1 mm and distance of select from screen is 1 m.Find minimum distance on screen from centre O where Maxima of both light coincide.​

Answer»

Here,

D = 1m

d = 1nm

\lambda_1 = 6000 \: A \degree \\ \\ \lambda_2 = 8000 \: A \degree

Let us consider that nth Maxima of

6000 A° and

mth maximum of 8000 A° coincides.

So,

\sf y_1 = y_2 \\ \\ \sf\frac{<klux>N</klux> \lambda_1 D }{d} = \frac{m \lambda_2 D }{d} \\ \\ \sf \frac{ \lambda_1}{ \lambda_2} = \frac{m}{n} \\ \\ \sf \purple{ \boxed{{\bold {m = \frac{<klux>3</klux>}{4} n}}}}

Put

n = 4 and m = 3

☆ 4th bright fringe of

6000A° and 3rd bright

fringe of 8000A° will be at

same place .

i.e.

Minimum distance from centre where the coinside

is

\sf \large \: y = \frac{n \lambda_1 D }{d} \\ \\ = \sf \frac{4 \times 6000 \times {10}^{ - 10} \times 1 }{ {10}^{ - 3} } \\ \\ \huge \green{ \bold{ \boxed { \boxed{ \sf y = 24 \times {10}^{ - 4} m}}}}


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