Question

determine the power of thin lens whose index of refraction is 1.50

Answer 1

For the lens in air

f

a

1

=(aμg−1)[

R

1

1

−

R

2

1

]

⇒

15

1

=(1.5−1)[

R

1

1

−

R

2

1

]

⇒

15

2

=

R

1

1

−

R

2

1

When the lens is immersed in water

f

w

1

=(

aμw

aμg

−1)[

R

1

1

−

R

2

1

]

⇒

f

w

1

=

⎝

⎜

⎜

⎛

3

4

1.5

−1

⎠

⎟

⎟

⎞

[

R

1

1

−

R

2

1

]

⇒

f

w

1

[

4

4.5

−1][

R

1

1

−

R

2

1

]

⇒

f

w

1

=

8

1

×

15

2

⇒

f

w

1

=

60

1

⇒f

w

=60cm.