In: Physics
Using what we know about the conjugate matrix, find the front principle plane for a thick lens with front and back radii of curvatures with a magnitude of 10.36 mm, a center thickness of 7.35 mm and an index of refraction of 1.45. What is the distance between the front surface of the lens and the principle plane in mm?
GIVEN: Refractive index of the lens, n = 1.45
The radius of curvature of front lens (let lens '1'), R1 = + 10.36 mm
The radius of curvature of back lens (let lens '2'), R2 = - 10.36 mm
(Positive and negative sign indicates the radii of curvature towards the right side and left side of the optical centre of the lens respectively, as shown in the image)
The thickness of the lens, d = 7.35 mm
TO FIND: The front principal plane and its distance from the front surface of the lens, h1 =?
SOLUTION: As shown in the image given below,
the distance between the principal plane of the front lens and front surface is given by,
....................(i)
here, f - focal length of the system of the thick lens and is given by lens maker's formula, i.e.,
................(ii)
Now, solving for 'f'. i.e,
Now, putting 'f' into equation (i) and solving for 'h1', we get
The required distance between the front surface and the front principal plane/ point and the front principal plane is drawn as shown in the image above.