Defects of Images and Vision Formulas

1. Chromatic aberration

The difference between f_v and f_R is a measure of longitudinal chromatic ) aberration, (L.C.A.) i.e.
L.C.A. = f_R – f_v with df = f_v – f_r
As, \(\frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
\(-\frac{\mathrm{df}}{\mathrm{f}^{2}}=\mathrm{d} \mu\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\)
\(-\frac{\mathrm{df}}{\mathrm{f}}=\frac{\mathrm{d} \mu}{\mu-1}=\omega\) as ω = \(\frac{\mathrm{d} \mu}{\mu-1}\)
L.C.A. = -df = ω f
Now as far as single lens neither f nor can be zero, we can not have a single lens free from chromatic aberration.

2. Condition of achromatism

\(\frac{\omega_{1}}{\mathrm{f}_{1}}+\frac{\omega_{2}}{\mathrm{f}_{2}}=0 \text { or } \frac{\omega_{1}}{\omega_{2}}=-\frac{\mathrm{f}_{1}}{\mathrm{f}_{2}}\)

 

3. Difect of vision

(i) Myopia or short sightedness
In this, distant objects are not clearly visible u = Γê₧, v = – x
power of lens P = \(\frac{1}{f}=\frac{1}{-x}\)
To overcome from this problem a concave lens is used.

(ii) Hypermetropia or long sightedness:
In this, near objects are not clearly visible
u = – 25 cm, v = – y
P = \(\frac{1}{f}=\frac{1}{25}-\frac{1}{y}\)
To overcome from this problem a convex lens is used.
(A)

(B)

(C)

Check out the go-to sheet for Formulas regarding various subjects like Physics, Chemistry, Maths all under one roof at Physicscalc.Com