Question

A lens that is "optically perfect" is still limited by diffraction effects. Suppose a lens has a diameter of 130 mm and a focal length of 680 mm . Part A: Find the angular width (that is, the angle from the bottom to the top) of the central maximum in the diffraction pattern formed by this lens when illuminated with 600
Part B: What is the linear width (diameter) of the central maximum at the focal distance of the lens?

Answer 1

Answer:

(A).  The angular width is $11.26\times10^{-6}\ rad$.

(B). The linear width is $7.64\times10^{-6}\ m$

Explanation:

Given that,

Diameter of lens = 130 mm

Focal length = 680 mm

Wavelength = 600 nm

(A). We need to calculate the angular radius of the central maxima

Using formula of the angular radius

$\theta =\dfrac{1.22\lambda}{D}$

Put the value into the formula

$\theta=\dfrac{1.22\times600\times10^{-9}}{0.13}$

$\theta=5.63\times10^{-6}\ rad$

We need to calculate the angular width

Using formula of angular width

$d = 2\theta$

Put the value into the formula

$d=2\times5.63\times10^{-6}$

$d=11.26\times10^{-6}\ rad$

(B). We need to calculate the radius of the central maximum at the focal distance of the lens

Using formula of radius

$R=\dfrac{1.22fd}{D}$

Put the value into the formula

$R=\dfrac{1.22\times0.68\times600\times10^{-9}}{0.13}$

$R=0.00000382 =3.82\times10^{-6}\ m$

We need to calculate the linear width

Using formula of linear width

$d=2R$

Put the value into the formula

$d=2\times3.82\times10^{-6}$

$d=0.00000764= 7.64\times10^{-6}\ m$

Hence, (A).  The angular width is $11.26\times10^{-6}\ rad$.

(B). The linear width is $7.64\times10^{-6}\ m$