Question

Two identical thin plano-convex glass lenses (refractive index 1.5) each having a radius of curvature of 20 cm are placed with their convex surfaces in contact at the center. The intervening space is filled with the oil of refractive index 1.7. The focal length of the combination is 

(A). -50 cm

(B). 50 cm

(C). -20 cm

(D). -25 cm

Answer 1

Hint: Solve the question by directly using the lens maker’s formula for the plano-convex lens, and then by keeping the values of the known parameters which are refractive index, the radius of curvature and then find the unknown values.

Formula used: ${\displaystyle \frac{1}{f}}=(\mu -1)\left({\displaystyle \frac{1}{R_1}}-{\displaystyle \frac{1}{R_2}}\right)$ , ${\displaystyle \frac{1}{f_{eq}}}=2\times {\displaystyle \frac{1}{f_{\text{lens}}}}+{\displaystyle \frac{1}{f_{\text{concave}}}}$

Complete step-by-step solution:

We have given that-

Refractive index of plano-convex glass lenses is ${\mu }_g=1.5$ 

Radius of the two lenses is R = 20 cm

Refractive index of oil is  ${\mu }_{oil}=1.7$

Now, we know from lens maker formula for plano-convex lens-

${\displaystyle \frac{1}{f}}=(\mu -1)\left({\displaystyle \frac{1}{R_1}}-{\displaystyle \frac{1}{R_2}}\right)$

Here, f is the focal length of the lens, $\mu$ is the refractive index.

Here R1 = R and for a plane surface $R_2=\mathrm{\infty }$ and ${\mu }_g=1.5$

${\displaystyle \frac{1}{f_{lens}}}=(1.5-1)\left({\displaystyle \frac{1}{R}}-0\right)={\displaystyle \frac{0.5}{R}}$

When the intervening medium is filled with oil then, it will act as a concave lens of refractive index 1.7 . 

${\displaystyle \frac{1}{f_{concave}}}=(1.7-1)\left(-{\displaystyle \frac{1}{R}}-{\displaystyle \frac{1}{R}}\right)=-0.7\times {\displaystyle \frac{2}{R}}={\displaystyle \frac{-1.4}{R}}$

Now, here we have two plano-convex lens and one  concave lens,

So,  ${\displaystyle \frac{1}{f_{eq}}}=2\times {\displaystyle \frac{1}{f_{lens}}}+{\displaystyle \frac{1}{f_{concave}}}$

Putting the values, we get-

${\displaystyle \frac{1}{f_{eq}}}=2\times {\displaystyle \frac{0.5}{R}}+{\displaystyle \frac{-1.4}{R}}={\displaystyle \frac{1}{R}}-{\displaystyle \frac{1.4}{R}}={\displaystyle \frac{-0.4}{R}}$

$\Rightarrow f_{eq}={\displaystyle \frac{-R}{0.4}}$

Now R = 20 cm given.

$\Rightarrow f_{eq}={\displaystyle \frac{-20}{0.4}}=-50$

Therefore, the focal length of the combination is -50 cm.

Hence, the correct option is (A).

 

Note: Plano-convex lenses have one spherical surface and one flat surface. The radius of the curvature of the plane surface is infinite. So, as mentioned in the solution by using a lens maker formula, find the focal length and then find the focal length of the combination.