Question

A spherical convex surface separates object and image spaces of refractive indices $1.0$ and $\dfrac{4}{3}$. If the radius of curvature of the surface is $10cm$, find its power.
a. $3.5D$
b. $2.5D$
c. $25D$
d. $1.5D$

Answer 1

A spherical convex surface is a part of a sphere which is rounded outwards. The point where all the incident parallel rays converge or diverge is called focus. Since power is to be calculated, so firstly focus is to be known. $f = \dfrac{{10(\dfrac{4}{3})}}{{\dfrac{4}{3} - 1}}$ or this suppose that the object is placed at point infinity and the image formed be at $f$.

Complete step by step answer:
Step 1:
Suppose $f$ be the focal point.
Given $R = 10$cm.
Refractive index of image, ${\mu _2}$= $\dfrac{4}{3}$
Refractive index of object, ${\mu _1}$= $1.0$
Position of object,$u = \infty$
Position of image, $v = f$

Step 2:
Using the relation
$\dfrac{{{\mu _2}}}{v}$-$\dfrac{{{\mu _1}}}{u}$=$\dfrac{{{\mu _2} - {\mu _1}}}{R}$ ……(i)

Step 3:
Substituting the values of 1, 2, v and u in equation (i),
$\dfrac{{{\mu _2}}}{f}$-$\dfrac{{{\mu _1}}}{\infty }$=2-1R

Step 4:
Cross Multiplying and evaluating value of $f$
$f = \dfrac{{R{\mu _2}}}{{{\mu _2} - {\mu _1}}}$
Substituting the values and solving for R,
$f = \dfrac{{10(\dfrac{4}{3})}}{{\dfrac{4}{3} - 1}}$ $f = \dfrac{{\dfrac{{40}}{3}}}{{\dfrac{1}{3}}}$
$f = \dfrac{{40}}{3}.\dfrac{3}{1}$
$f = 40cm$
Convert cm to m, and evaluating value of f
$f = \dfrac{{40}}{{100}} = 0.4m$

Step 5:
Since power is the inverse of focal length, so it’s formula can be written as
$P = \dfrac{1}{f}$
$\Rightarrow \dfrac{1}{{0.4}}$
$\Rightarrow 0.25D$
The power of a spherical convex surface is $2.5D$.

Hence, the correct answer is option (B).

Note: From above it can be noticed that the power of a lens is inversely proportional to its focal length. If in a convex lens focal length is small then the power will be more and hence it will be more powerful. On the other hand, if focal length is large the power will be small and the lens will be less powerful. One property of convex lenses is that they always have a positive focal length. Since the focal length is always positive therefore they always have positive values of power and hence power of a convex lens can not be negative.