Question

Q. 6: Design a single lens collimator to be used with a He-Ne laser (wavelength 632.8nm). Focal length of the collimator is 50cm. What are the radii of the two surfaces for the spherical aberration to be minimum. Use BK7 glass.

Answer 1

For BK7 glass, the refractive index at 632.8 nm (He-Ne laser) is approximately,

n = 1.515

Let the radii of curvature of the lenses be r₁ and r₂

The focal length of the lens is given by the formula

- (-1) |

Theoretically, the minimum spherical aberration occurs when

71_n(2n – 1) - 4 n(2n +1)

Using n = 1.515,

= (1.515+ (2+1.515-11-4)/(1.515*(2*1.515+1)) = -0.1514302795

So,

r = -0.1514302795 *

Using the lens equation, with f = 50 cm = 0.5 m

= (0.515) * :-0.1514302795 - 1/2

(- sign means the direction of curvature)

r2 = (0.515)* [- -0.1514302795–10.5 = -1.9579m

And

r1 = -0.1514302795 * -1.9579 = 0.29648m