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Brums [2.3K]
3 years ago
14

Light of wavelength 597 nm falls on a double slit, and the first bright fringe of the interference pattern is seen at an angle o

f 15.8° with the horizontal. Find the separation between the slits. µm
Physics
2 answers:
Kazeer [188]3 years ago
8 0

Answer:

2.2 µm

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.

Given wavelength = 597 nm

Angle, \theta  = 15.8°

First bright fringe means , m = 1

So,

d\times sin\ 15.8^0=1\times \597\ nm

d\times 0.2723=1\times \597\ nm

d=2192.43481\ nm

Also,

1 nm = 10⁻⁹ m

1 µm = 10⁻⁶ m

So,

1 nm = 10⁻³ nm

Thus,

<u>Distance between slits ≅ 2.2 µm</u>

yarga [219]3 years ago
4 0

Answer:

The separation between the slit is 2.19\mu m

Solution:

As per the question:

Wavelength of light, \lambda = 597 nm = 597\times 10^{-9} m

\theta = 15.8^{\circ}

Now, by Young's double slit experiment:

xsin\theta = n\lambda

here,

n = 1

x = slit width

Therefore,

x = \frac{597\times 10^{-9}}{sin15.8^{\circ}} = 2.19\times 10^{- 6} m

x = 2.19\mu m

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