Question

Find the distance in nm between two slits that produces the first minimum for 410-nm violet light at an angle of 14.5°.

Answer 1

Answer:

820 nm

Explanation:

We are given that

Wavelength=$\lambda=410 nm$

$\lambda=410\times 10^{-9} m$

$1nm=10^{-9} m$

$\theta=14.5^{\circ}$

For first minimum therefore

m=0

We know that for destructive interference

$(m+\frac{1}{2})\lambda=dsin\theta$

Substitute the values

$(0+\frac{1}{2})\times 410\times 10^{-9}=dsin 14.5$

$d=\frac{410\times 10^{-9}}{2\times sin 14.5}$

$d=820\times 10^{-9} m=820 nm$

Hence, the distance between two slits that produces the first minimum=820 nm

Answer 2

Answer:

Explanation:

wavelength, λ = 410 nm

Angle, θ = 14.5°

The formula used for the distance between the two slits for minima is given by

$dSin\theta =\left (m+\frac{1}{2} \right )\lambda$

for first order minima, m = 0

d Sin 14.5 = 0.5 x 410

d = 818.76 nm