# Two identical thin Plano- convex glass lenses (refractive index $1.5$) each having a radius of curvature of $20\;{\text{cm}}$ are placed with their convex surfaces in contact at the center. The intervening space is filled with oil of refractive index $1.7$. The local length of the combination is :

(A) $ - 20\;{\text{cm}}$

(B) $ - 25\;{\text{cm}}$

(C) $ - 50\;{\text{cm}}$

(D) $50\;{\text{cm}}$

Answer

Verified

290.4k+ 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.

Last updated date: 31st May 2023

•

Total views: 290.4k

•

Views today: 4.46k

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Name the Largest and the Smallest Cell in the Human Body ?

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

A ball impinges directly on a similar ball at rest class 11 physics CBSE

Lysosomes are known as suicidal bags of cell why class 11 biology CBSE

How do you define least count for Vernier Calipers class 12 physics CBSE

Two balls are dropped from different heights at different class 11 physics CBSE

A 30 solution of H2O2 is marketed as 100 volume hydrogen class 11 chemistry JEE_Main