Question

Q. Convex lens made up of glass of refractive index 1.5 is dipped in turn in
(i) Medium A of n=1.65 ii) Medium B of n = 1.33.
Explain giving reasons Whether it will behave as a converging lens or diverging lens in each of the

Answer 1

Answer:Two cases.

Here ng

Let fair be the focal length of lens in air then

1/fair = (ng-1)(1/R1-1/R2)

(1/R1-1/R2) = 1/fair(ng-1) = 2/fair……………………………..(1)

(i) When lens is dipped in a medium A here nA = 1.65

Focal length be fA when dipped in a medium A then 1/fA=(nA-1)(1/R1-1/R2)

Using equation (1) we have

1/fA = (1.5/1.6 5-1) X 2/fair = -1/5.5 fair

FA = -5.5 fair

As sign of fA is opposite to that of fair, the lens will behave as diverging lense.

 

(ii) When lens is dipped in a Medium B nB = 1.33

Let fB be the focal length of lens when dipped in medium B

Then 1/fB = (ng-1)(1/R1-1/R2) =(ng/nB-1) )(1/R1-1/R2)

1/fB =(1.5/1.33-1)X2/fair = 0.34/1.33 fair = FB = 3.91

fair as the Sign of fB is same as that of fair the lens will behave as a converging lens

Q. A convergent beam of light passes through a diverging lens of focal length 0.2 meters comes to focus at a distance 0.3 meters behind the lens find the position of the point at which the beam would converge in the absence of lens?

Answer: F=-0.2 m, v=0.3 m

From the lens equation 1/f = 1/v-1/u

1/u = 1/v-1/f = 1/0.3-1/-0.2 = 50/6

 

U=6/50 = 0.12 m

In the absence of lens the beam would converge at a distance 0.12 m from the present position of the lens.

Q. A beam of light converges to a point P. A lens is placed in the path of convergent beam 12 cm from the point P. At what point the beam converges if the lens is (a) a concave lens of focal length 16 cm (b) a convex length of focal length 20 cm.

Answer: (a) here u =12cm f = -16 cm the lens equation we have

V=uf/u+f=12X-16/12-16 = 48 cm  

as v is positive the beam converges on the same side that of point P

b) Here u = 12 cm , f =20 cm from the lens equation we have

V=uf/u+f = 12X20/12+20 = 240/32 = 7.5 cm

As v is positive the beam converges on the same side as that of point of P

Q.A converging and a diverging of equal focal lengths are placed co-axially in Contact. Find the focal length and power of the combination.

Answer: Let f and –f are be the focal length of the converging and diverging lens respectively then focal length of the combination

1/F = 1/f – 1/f = 0

Power of the combination p = 1/F = 0.

Explanation:

Answer 2

Answer:

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Explanation: