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Aleksandr [31]
3 years ago
5

What is the width of a slit for which the first minimum is at 45° when the slit is illuminated by a helium-neon laser (λ = 633 n

m)?
Hint: The small-angle approximation is not valid at 45°.
Physics
1 answer:
bekas [8.4K]3 years ago
4 0

Answer:

d = 895 nm

Explanation:

Given:

- Wavelength of light λ = 633 nm

- Angle Q is between central order = 45 degrees

Find:

What is the width of a slit?

Solution:

- The relationship between and wavelength and the order of fringes with respect to slit size d is given by Young's experiment as follows:

                                    sin (Q) = m*λ / d

- Use the above result and compute for the grating d for m =1,

                                     d = λ / sin(Q)

                                     d = 633*10^-9 / sin(45)

                                    d = 0.89 um

-  The width of slit is d = 895 nm

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