Question

Lenses are constructed by a material of refractive index 1.50. The magnitude of the radii of curvature are 20 cm and 30 cm. Find the focal lengths of the possible lenses with the above specifications.

Answer 1

Given :

The possibel cases are as follows:

R1=20cm

'R2= 30cm

n=1.5

Lens makers fromula :

1/f= (n-1) [1/R1-1/R2]

For Double Convex lens :

R1=+20cm R2= - 30cm

1/f= (1.5-1)[1/20+1/30]

=0.5 x [ 3+2/60]

=0.5 x 5/60

= 5x5/ 10 x 60

=1/24

f= 24cm

For Double concave lens :

R1= - 20cm R2=+30cm

1/f =(1.5-1)[-1/20 -1/30]

=0.5x [ -3-2/60]

f= -24cm

For Concave Convex :

R1= +20cm

R2= +30

1/f= 0.5[ 1/20-1/30]

= 0.5 x [ 3-2/60]

=5/10x60

f= 120cm

For concave concave lens :

R1= -20cm

R2= - 30cm

1/f = 0.5x[ -1/20+1/30]

= 0.5 [ -3+2/60]

f= -1 20cm

Answer 2

$\huge{\mathfrak{\blue{ELLO}}}$

==>>

Refer attachment.........

respective focal lengths are..

F1 = +24cm(double convex)
F2= -24cm(double concave)
F3= 120cm(concave concave)
F4= -120cm(concave convex)...

$\huge{\underline{\mathfrak{\red{it's\:@\:styoo}}}}$