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Yakvenalex [24]
3 years ago
13

Green light (λ = 518 nm) strikes a single slit at normal incidence. What width slit will produce a central maximum that is 3.00

cm wide on a screen 2.00 m from the slit? Answer in μm .
Physics
1 answer:
MariettaO [177]3 years ago
7 0

Answer:

6.9066 × 10⁻⁵ m

Explanation:

For constructive interference, the expression is:

d\times sin\theta=m\times \lambda

Where, m = 1, 2, .....

d is the distance between the slits.

The formula can be written as:

sin\theta=\frac {\lambda}{d}\times m ....1

The location of the bright fringe is determined by :

y=L\times tan\theta

Where, L is the distance between the slit and the screen.

For small angle , sin\theta=tan\theta

So,  

Formula becomes:

y=L\times sin\theta

Using 1, we get:

y=L\times \frac {\lambda}{d}\times m

Thus, the distance between the central maximum is 3.00 cm

First bright fringe , m = 1 occur at 3.00 / 2 = 1.50 cm

Since,

1 cm = 0.01 m

y = 0.0150 m

Given L = 2.00 m

λ = 518 nm

Since, 1 nm = 10⁻⁹ m

So,

λ = 518 × 10⁻⁹ m

Applying the formula as:

0.0150\ m=2.00\ m\times \frac {518\times 10^{-9}\ m}{d}\times 1

<u>⇒ d, distance between the slits = 6.9066 × 10⁻⁵ m</u>

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