Answer

Verified

53.4k+ views

**Hint**The refractive index of the lens and the radius of curvature are two factors affecting the focal length of the lenses. Lens maker’s formula can relate the dependency easily. It is given as

$\dfrac{1}{f} = \left( {\mu - 1} \right)\left[ {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right]$

Where $\mu $is the refractive index and${R_1}$, ${R_2}$are the radius of curvature of the surfaces.

And for the plane surfaces the radius of curvature would be infinity.

**Complete Step by step solution**

We have given two identical thin Plano convex glass lenses.

Refractive index of the Plano convex glass lenses, $\mu = 1.5$

Radius of curvature, $R = 20\;{\text{cm}}$

Refractive index of the oil, $\mu ' = 1.7$

For the Plano convex glass lenses, the Lens maker's formula is given as,

$\dfrac{1}{f} = \left( {\mu - 1} \right)\left[ {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right]$

Where $\mu $is the refractive index and${R_1}$,${R_2}$are the radius of curvature of the surfaces.

Here, ${R_1} = R$and the next surface is plane. Therefore, ${R_2} = \alpha $. For the plane surfaces the radius of curvature is infinity.

Therefore substituting those in the above expression gives,

$

\dfrac{1}{f} = \left( {1.5 - 1} \right)\left[ {\dfrac{1}{R} - \dfrac{1}{\alpha }} \right] \\

= 0.5\left[ {\dfrac{1}{R} - 0} \right] \\

= \dfrac{{0.5}}{R} \\

$

The intervening medium is formed like the concave lens. It was filled with the oil.

Using the Lens maker's formula for the concave lens,

$\dfrac{1}{{f'}} = \left( {\mu ' - 1} \right)\left[ { - \dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right]$

Here both ${R_1}$and ${R_2}$are $R$

Therefore,

$

\dfrac{1}{{f'}} = \left( {1.7 - 1} \right)\left[ { - \dfrac{1}{R} - \dfrac{1}{R}} \right] \\

= 0.7 \times - \dfrac{2}{R} \\

= - \dfrac{{1.4}}{R} \\

$

The combination includes two identical Plano convex lenses and the intervening portion like convex lens. The combined focal length is given as,

$

\dfrac{1}{{{f_c}}} = \dfrac{1}{f} + \dfrac{1}{f} + \dfrac{1}{{f'}} \\

= 2 \times \dfrac{1}{f} + \dfrac{1}{{f'}} \\

$

Substituting the values in the above expression,

$

\dfrac{1}{{{f_c}}} = 2 \times \dfrac{1}{f} + \dfrac{1}{{f'}} \\

= 2 \times \dfrac{{0.5}}{R} + \dfrac{{ - 1.4}}{R} \\

= - \dfrac{{0.4}}{R} \\

$

Therefore,

${f_c} = - \dfrac{R}{{0.4}}$

Substitute the value of radius of curvature.

$

{f_c} = - \dfrac{{20\;{\text{cm}}}}{{0.4}} \\

= - 50\;{\text{cm}} \\

$

Thus the combined focal length is $ - 50\;{\text{cm}}$.

**Thus the answer is Option C.**

**Note**while using the Lens maker’s formula certain sign conventions have to be noted. The convex lens is a converging lens and the focal length will be positive. Then, in the Lens maker’s formula

$\dfrac{1}{f} = \left( {\mu - 1} \right)\left[ {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right]$

${R_1}$ is positive and ${R_2}$is negative.

The concave lens is a diverging lens and the focal length is negative. Then,

$\dfrac{1}{f} = \left( {\mu - 1} \right)\left[ { - \dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right]$

${R_1}$ is negative and ${R_2}$is positive.

Recently Updated Pages

Why do we see inverted images in a spoon class 10 physics JEE_Main

A farsighted man who has lost his spectacles reads class 10 physics JEE_Main

State whether true or false The outermost layer of class 10 physics JEE_Main

Which is not the correct advantage of parallel combination class 10 physics JEE_Main

State two factors upon which the heat absorbed by a class 10 physics JEE_Main

What will be the halflife of a first order reaction class 12 chemistry JEE_Main

Other Pages

The nitride ion in lithium nitride is composed of A class 11 chemistry JEE_Main

The resultant of vec A and vec B is perpendicular to class 11 physics JEE_Main

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

A count rate meter is used to measure the activity class 11 chemistry JEE_Main

Electric field due to uniformly charged sphere class 12 physics JEE_Main

when an object Is placed at a distance of 60 cm from class 12 physics JEE_Main