Optics Tutorial 5 HW solutions

1. A customer calls in and is using an RKE eyepiece (Edmund Optics part number 30941) with a focal length of 21.5 mm. He places an LCD at perfect focus using an autocollimator. When he looks at it he has to move the LCD 0.123 mm closer to the eyepiece. After asking – you find out that the customer is not wearing glasses when making the observation. What is the power of his eyeglass prescription? 

Solution: One could start with Newton's thin lens equation. Rather I'll use the equation from slide 15 that converts a linear refocus to power:

φ (diopters) = - x/F^2 * 1000 , where x is the linear defocus in mm
φ (diopters) = -( -0.123)/21.5^2 * 1000 = 0.266 diopters
Lens Crafters(TM) makes him glasses with ~1/4 diopter correction

2. What is the radius of curvature of the wavefront at the entrance to the eyepiece (between the customers eye and the eyepiece?

Solution: A quarter diopter is 0.25/m. Invert this to get the radius of the wavefront or ... 4 meters.
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