Question

if the near point N= 21.2 cm, the eyepiece focal length f eye=5.07mm, the distance between lenese D=21.07 and total magnification of M=407 what is the absolute valye of the power of the objective lens

Answer 1

The absolute value of the objective lens power is approximately 326.67 diopters.

What is the absolute value of the power of the objective lens?

The power of a lens (P) is the reciprocal of its focal length (f) in meters:

P = 1 / f

Convert the near point and eyepiece focal length to meters:

N = 21.2 cm = 0.212 m

f eye = 5.07 mm = 0.00507 m

We can use the formula for total magnification (M) and rearrange it to solve for f obj:

M = - (D + N * f obj) / (f obj * f_eye)

f_obj = (D + N * M) / (M * f_eye - N)

f obj = (0.2107 + 0.212 * 407) / (407 * 0.00507 - 0.212)

f obj ≈ 0.00306 m

P obj = 1 / f obj ≈ 326.67 diopters (D)

Finally, take the absolute value of the power:

|P obj| ≈ 326.67 D

Therefore, the absolute value of the objective lens power is approximately 326.67 diopters.