(A) (i) Define the S.I. unit of power of a lens. (ii) A plano-convex lens is made of glass of refractive index $1.5$. The radius of curvature of the convex surface is $25\ \mathrm{cm}$. (ii.i) Calculate the focal length of the lens. (ii.ii) If an object is placed $50\ \mathrm{cm}$ in front of the lens, find the nature and position of the image formed.

(i) One dioptre (D) is the power of a lens of focal length $1\ \mathrm{m}$. (ii.i) $f=+50\ \mathrm{cm}$. (ii.ii) The object is at the focus, so the emergent rays are parallel and the image forms at infinity.

CBSE Class 12 · 2022 · Geometrical OpticsShort Answer

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(A) (i) Define the S.I. unit of power of a lens. (ii) A plano-convex lens is made of glass of refractive index

1.5

. The radius of curvature of the convex surface is

25 cm

. (ii.i) Calculate the focal length of the lens. (ii.ii) If an object is placed

50 cm

in front of the lens, find the nature and position of the image formed.

Solution

(i) \textbf{Dioptre (D):} the power of a lens whose focal length is

1 metre

(P = \frac{1}{f ( in m )})

.
(ii.i) Plano-convex (

R_{1} = 25 cm

R_{2} = \infty

\frac{1}{f} = (μ - 1) \frac{1}{R} = (1.5 - 1) \frac{1}{25} = \frac{1}{50} \Rightarrow f = + 50 cm

(ii.ii) With

u = - 50 cm = - f

, the object lies at the focus:

\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{50} - \frac{1}{50} = 0 \Rightarrow v \to \infty

Image at infinity (emergent rays parallel)

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CBSE Class 12 · 2022 · Geometrical OpticsShort Answer

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