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# 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) $50\;{\text{cm}}$

Last updated date: 18th Jun 2024
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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.