Question

What must be the ratio of the slit width to the wavelength for a single slit to have the first diffraction minimum at 45.0°?

Answer 1

Given that,

Angle of first diffraction = 45°C

We know that,

The distance between two slit is

$d\sin\theta=n\lambda$

Where, d = distance

$\lambda$ = wavelength

$\theta$ = angle

For first minimum,

n = 1

$d\sin\theta=\lambda$

We need to calculate the ratio of the slit width to the wavelength for a single slit

Using formula of distance between two slit

$\dfrac{d}{\lambda}=\dfrac{1}{\sin\theta}$

Put the value into the formula

$\dfrac{d}{\lambda}=\dfrac{1}{\sin45}$

$\dfrac{d}{\lambda}=1.414$

Hence, The ratio of the slit width to the wavelength for a single slit is 1.414