Question

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Answer 1

Questions—

In YDSE distance between slits and screen is 1.5m.When light of wavelength 500nm is used then 2nd bright fringe is obtained on screen at a distance of 10mm from central bright fringe.What will be shift in position of 2nd bright fringe if light of wavelength 550nm is used.

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$\huge\sf\underline{Answer}$

$\bf\huge\boxed{1mm}$

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Given—

Initially D = 1.5 m

λ = 500 nm = 500 × 10^(-9) m

n = 2

Y2= 10 mm = 10¯² m

We know

In YDSE,the formula for bright fringes (constructive interference) is:

Solution—

$\huge Yn = \frac{nλD}{d}$

for n=2

10¯² = {2× 500× 10^(-9) × 1.5}/d

Solving We get

d= 1.5 × 10¯⁴ m

NOW if λ = 550 nm

Again YDSE

Y2' = {2×550×10^(-9)× 1.5} /{1.5×10¯⁴}

We get

Y2'= 1.1 × 10¯²

SHIFT—

Y2' - Y2 = 1.1× 10¯² - 1

=>10¯²(1.1-1)

=> 0.1 × 10¯² m = 1mm

Therefore, the shift in position of 2nd bright fringe if light of wavelength 550nm is 1 mm

Option(2) is correct

$\huge{\red{\ddot{\smile}}}$

Answer 2

Answer:

1 mm

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