Question 80: Ray Optics

Question

The magnitudes of power of a biconvex lens (refractive index 1.5) and that of a plano-concave lens (refractive index $=1.7$ ) are same. If the curvature of planoconcave lens exactly matches with the curvature of back surface of the biconvex lens, then ratio of radius of curvature of front and back surface of the biconvex lens is $\_\_\_\_$

Accepted Answer

If the focal length of a lens is $f$ (in meters) then its power is given as $P=\frac{1}{f} \ldots$ (i)
The Lens Maker's formula is given as $\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}ight)$, where $\mu$ is the refractive index of lens with respect to the medium, $R_1$ and $R_2$ are the radii of curvatures (measured using sign convention).
Using the Lens Maker's Formula and power formula :
$$ \mathrm{P}=\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}ight) $$
Here, $\mu_1=$ refractive index of biconvex lens, $R_1 \& R_2$ are the radii of curvatures of front and back curvature of biconvex lens.
Then power of biconvex lens is,
$$ P_{	ext {convex }}=(1.5-1)\left(\frac{1}{R_1}-\left(\frac{1}{-R_2}ight)ight)=0.5\left(\frac{1}{R_1}+\frac{1}{R_2}ight) $$
For the Plano-Concave Lens
Refractive Index $\mu_2=1.7$
The curvature of the plano-concave lens matches the back surface of the biconvex lens. Thus, its curved surface has radius $R_2$.
For a plano-concave lens, one surface is flat $R_3=\infty$ and the other is concave (radius $R_4=R_2$ ).
So, the power of Plano-concave lens is,
$$ P_{	ext {concave }}=(1.7-1)\left(\frac{1}{R_3}-\frac{1}{R_4}ight)=0.7\left(\frac{1}{\infty}-\frac{1}{R_2}ight)=-\frac{0.7}{R_2} $$
The magnitudes of power are the same
$$ \left|P_{	ext {convex }}ight|=\left|P_{	ext {concave }}ight| $$
$$\Rightarrow $$ $0.5\left(\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_2}ight)=\frac{0.7}{\mathrm{R}_2}$
$$\Rightarrow $$ $\frac{5}{\mathrm{R}_1}=\frac{7}{\mathrm{R}_2}-\frac{5}{\mathrm{R}_2}$
$$\Rightarrow $$ $\frac{5}{\mathrm{R}_1}=\frac{2}{\mathrm{R}_2}$
$$\Rightarrow $$ $\frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{5}{2}$
Here, $R_1 $ & $ R_2$ are the radii of curvatures of front and back curvatures whose ratio is $5: 2$.
Hence, the correct option is (B).

Answer

$2 : 5$

Answer

$5: 2$

Answer

$12: 5$

Answer

$5: 12$

Ray Optics

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