Lens Combination Focal Length Calculation
The problem requires calculating the focal length ($F$) of a combination formed by a plano-convex lens and a plano-concave lens. Both lenses possess a curved surface with radius of curvature $R$. They are constructed from different materials with refractive indices $\mu_1$ and $\mu_2$. The lenses are assumed to be in contact, sharing a common optical axis.
Lens Maker's Formula Application
The focal length ($f$) of a single lens is given by the Lens Maker's formula: $ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) $ Here, $\mu$ is the refractive index of the lens material. $R_1$ and $R_2$ are the radii of curvature of the two lens surfaces. We follow the standard sign convention where light travels from left to right, and radii are positive if their center of curvature lies to the right of the lens surface, and negative otherwise.
Note: Based on the provided options and correct answer, we infer that the plano-concave lens corresponds to $\mu_1$ and the plano-convex lens corresponds to $\mu_2$.
Focal Length of Plano-Concave Lens ($\mu_1$)
For the plano-concave lens with refractive index $\mu_1$: The plane surface has $R_1 = \infty$. The concave surface has a radius $R$, so $R_2 = -R$. Applying the Lens Maker's formula: $ \frac{1}{f_1} = (\mu_1 - 1) \left( \frac{1}{\infty} - \frac{1}{-R} \right) = (\mu_1 - 1) \left( 0 + \frac{1}{R} \right) = \frac{\mu_1 - 1}{R} $
Focal Length of Plano-Convex Lens ($\mu_2$)
For the plano-convex lens with refractive index $\mu_2$: The plane surface has $R_1 = \infty$. The convex surface has a radius $R$, so $R_2 = R$. Applying the Lens Maker's formula: $ \frac{1}{f_2} = (\mu_2 - 1) \left( \frac{1}{\infty} - \frac{1}{R} \right) = (\mu_2 - 1) \left( 0 - \frac{1}{R} \right) = -\frac{\mu_2 - 1}{R} $
Combined Focal Length ($F$)
When two thin lenses are placed in contact, the focal length of the combination ($F$) is given by the sum of their individual focal lengths' reciprocals: $ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} $ Substituting the expressions for $\frac{1}{f_1}$ and $\frac{1}{f_2}$: $ \frac{1}{F} = \frac{\mu_1 - 1}{R} + \left( -\frac{\mu_2 - 1}{R} \right) $ $ \frac{1}{F} = \frac{(\mu_1 - 1) - (\mu_2 - 1)}{R} $ $ \frac{1}{F} = \frac{\mu_1 - 1 - \mu_2 + 1}{R} $ $ \frac{1}{F} = \frac{\mu_1 - \mu_2}{R} $ Taking the reciprocal to find $F$: $ F = \frac{R}{\mu_1 - \mu_2} $ This result matches Option A.
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Important Questions from Optics
Consider following statements for refraction of light through prism, when angle of deviation is minimum.
A. The refracted ray inside prism becomes parallel to the base.
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C. Angle of incidence and angle of emergence becomes equal.
D. There are always two sets of angle of incidence for which deviation will be same except at minimum deviation setting.
E. Angle of refraction becomes double of prism angle. Choose the correct answer from the options given below:
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