Solved Problems on Optics
(by Dr. Ram Chand)
Problem 19
Two thin lenses have focal lengths of −5 and +20.
Determine their equivalent focal lengths when
(a)cemented together and
(b)separated by 10 cm.
Solution:
Given: f1 = −5cm and f2 = +20cm.
(a) Focal length when cemented together: for this we use
1
1
−3
+ 20
= −4+1
the formula: f1 = f11 + f12 = −5
20 = 20
f = − 20
3 = −6.7cm.
(b) Focal length when two lenses are 10cm apart. We know
that for the first lens if s01 = ∞, then s1 = f1 = −5cm.
Now this image will become an object for th second lens
placed at 10cm away from the first lens.
Negative s0 implies that the image is on the object side of
the first lens. Thus the object distance for the second lens
will be s2 = 5 + 10 = 15cm. Now using the following
equation: s12 + s10 = f12
2
1
s02
s01
1
1
= 20
− 15
= 3−4
60
= −60cm.
Thus the focal length of this system is f = −60cm.
Problem 24
A convex thin lens with refractive index of 1.50 has
a focal length of 30 cm in
air. When immersed in a
certain transparent liquid,
it becomes a negative lens
with a focal length of 188
cm. Determine the refractive index of the liquid.
Solution:
Refractive index of lens = n1 = 1.5
Focal length of the lens in air = f1 = 30cm
Focal length of the lens when immersed in the liquid of
refractive index, nL , f2 = −188cm
1
Using lensmaker formula: f1 = n2n−n
( R11 − R12 ).
1
1
1
1
= 1.5−1
In air: 30
1 ( R1 − R2 )
1
1
1
L
In liquid of refractive index nL : −188
= 1.5−n
nL ( R 1 − R 2 )
30
L
Diving first equation by 2nd: −188
= n1.5−n
L ×0.5
30 × 0.5nL , or 15nL = −188(1.5 − nL ) = −282 + 188nL
282
188nL − 15nL = 282, or nL = 173
= 1.63.
1