A glass convering questions for physics 2

Question

A glass converging lens (n=1.50) is designed to look like the lens.

The radius of the first surface is 15 cm and the radius of the second surface is 10 cm. a)find the focal lenght of the lens.

Determine the position of the image for object distances of b)infinity

c)3f d)f and e)f/2

Answer 1

For a lens (approximately)

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

R1 - radius of the frontal curvature (first surface)

R2- Radius of the posterior curvature

f- focal length

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

f=1/[(1.5 -1 ) ( 1/0.15 - 1/0.10)]

f= -0.6 m

also we know that

S2/S1= f/(f-S1)

S1 - distance of object from image to lens axis

S2 - distance of image from image to lens axis

then

S2= fS1/(f  -S1)

S2= f(f/S1 - 1)

b) S1= infinity

S2= f/(-1)= -f

c)S1= 3f

S2=3 f^2/(f - 3f)= 3/2f

e) S1= 0.5 f^2/( f-0.5 f)= f