1. \(\frac{1}{f}=\Big(\frac{n_2}{n_1}-1\Big)\Big(\frac{1}{R_1}-\frac{1}{R_2}\Big)\)
2. Refraction by a lens:
Lens Maker’s Formula (for a thin lens): Consider a thin lens of refractive index n2 formed by the spherical surfaces ABC and ADC. Let the lens is kept in a medium of refractive index n1 . Let an object ‘O’ is placed in the medium of refractive index n1 . Hence the incident ray OM is in the medium of refractive index n1 and the refracted ray MN is in the medium of refractive index n2 .
The spherical surface ABC (radius of curvature R,) forms the image at l1 . Let ‘u’ be the object distance and ‘v1 ’ be the image distance.
Then we can write,
\(\frac{n_2}{v_1}-\frac{n_1}{u}=\frac{n_2-n_1}{R_1}\) ......(1)
Adding eq (1) and eq (2) we get
\(\frac{n_2}{v_1}-\frac{n_1}{u}+\frac{n_1}{v}-\frac{n_2}{v_1}\) = \(\frac{n_2-n_1}{R_1}+\frac{n_1-n_2}{R_2}\)
\(\frac{n_1}{v}-\frac{n_1}{u}=(n_2-n_1)\)\(\Big(\frac{1}{R_1}-\frac{1}{R_2}\Big)\)
Dividing throughout by n1 , we get
\(\frac{1}{v}-\frac{1}{u}=\Big(\frac{n_2}{n_1}-1\Big)\)\(\Big(\frac{1}{R_1}-\frac{1}{R_2}\Big)\)
if the lens is kept in air ,\(\frac{n_2}{n_1}\) = n
So the above equation can be written as,
\(\frac{1}{v}-\frac{1}{u}\)\((n-1)\Big(\frac{1}{R_1}-\frac{1}{R_2}\Big)\) ......(4)
From the definition of the lens, we can take, when u = ∞, f = v
Substituting these values in the eq (3), we get
∴ \(\frac{1}{f}-\frac{1}{∞}\)\(=(n-1)\Big(\frac{1}{R_1}-\frac{1}{R_2}\Big)\)
This is lens maker’s formula
\(\frac{1}{f}=(n-1)\Big(\frac{1}{R_1}-\frac{1}{R_2}\Big)\) ......(5)
For convex lens.
f = +ve, R = +ve, R = – ve
\(\frac{1}{f}=(n-1)\Big(\frac{1}{R_1}+\frac{1}{R_2}\Big)\)
For concave lens,
f = -ve, R = -ve, R = +ve
\(\frac{1}{f}=(n-1)\Big(\frac{1}{R_1}-\frac{1}{R_2}\Big)\)
Lens formula
From eq(4),
\(\frac{1}{v}-\frac{1}{u}=(n-1)\Big(\frac{1}{R_1}-\frac{1}{R_2}\Big)\)
From eq(5)
\(\frac{1}{f}=(n-1)\Big(\frac{1}{R_1}-\frac{1}{R_2}\Big)\)
From these two equations, we get
\(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
Linear magnification :
If h is the height of the object and ho is the height of the image, then linear magnification.
\(m=\frac{h_1}{h_o}=\frac{v}{u}\)
3. a. R1 = R, R2 = +R
∴ \(\frac{1}{f}=\Big(\frac{n_2}{n_1}-1\Big)\Big(\frac{1}{R}-\frac{1}{R}\Big)\)
power of lens, P = 0
b. We know
\(\frac{1}{f}=\Big(\frac{n_2}{n_1}-1\Big)\Big(\frac{1}{R_1}-\frac{1}{R_2}\Big)\)
The above equation shows when n1 increases f decreases the refractive index of water is greater than air. Hence when we place a lens in water, focul length decreases.