Question

A single slit forms a diffraction pattern, with the first minimum at an angle of 40.0° from central maximum, when monochromatic light of 630-nm wavelength is used. The same slit, illuminated by a new monochromatic light source, produces a diffraction pattern with the second minimum at a 60.0° angle from the central maximum. What is the wavelength of this new light?

Answer 1

Answer:

$\lambda = 424.4 nm$

Explanation:

As we know by the formula of diffraction

$a sin\theta = N\lambda$

so we have

a = slit size

$\theta$ = angular position of Nth minimum

so we will have

for first minimum of 630 nm light

$a sin40 = 1(630 \times 10^{-9})$

$a = 9.8 \times 10^{-7} m$

Now for another wavelength second minimum is at 60 degree angle

$a sin60 = 2 \lambda$

$(9.8 \times 10^{-7}) sin60 = 2 \lambda$

$\lambda = 424.4 nm$