Question

a biconvex LENS R. I.=1.5 has equal curvatureeach of 20 CM the power of lens is what GIVE SOLUTION

Answer 1

Use the formula,
$\bold{\frac{1}{f}=(\mu-1)[\frac{1}{R_1}-\frac{1}{R_2}}]$
here, μ = 1.5
R₁ = 20cm and R₂ = -20 cm
Then, 1/f = (1.5 -1)[ 1/20 - 1/-20] = 1/20
hence, f = 20cm = 0.2m

Now, power ( P ) = 1/focal length of lens = 1/f
Power = 1/0.2 = 5D

Answer 2

Solution:
n=1.5
R1=R2=20cm
P=?

use the Lens makers Formula :
1/f=(n-1)(1/R1-1R2)

For convex lens R=+20cm and R2=-20cm

1/f=(1.5-1)[1/20+1/20]
=0.5 x2/20
=5x2/10 x20
=1/20
1/f=1/20cm
f=20cm=20/100=0.2m
As we Know that :
P=1/f
=1/0.2
=10/2
=5D