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
- In a microscope the objective is having focal length $f_o = 2 \text{ cm}$ and eye-piece is having focal length $f_e = 4 \text{ cm}$. The tube length is $32 \text{ cm}$. The magnification produced by this microscope for normal adjustment is ____________.
- A collimated beam of light of diameter $2 \text{ mm}$ is propagating along x-axis. The beam is required to be expanded in a collimated beam of diameter $14 \text{ mm}$ using a system of two convex lenses. If first lens has focal length $40 \text{ mm}$, then the focal length of second lens is ____________ $\text{mm}$.
- In a Young's double slit experiment set up, the two slits are kept $0.4\text{ mm}$ apart and screen is placed at $1\text{ m}$ from slits. If a thin transparent sheet of thickness $20 \, \mu\text{m}$ is introduced in front of one of the slits then center bright fringe shifts by $20\text{ mm}$ on the screen. The refractive index of transparent sheet is given by $\frac{\alpha}{10}$, where $\alpha$ is __________.
- In two separate Young's double-slit experimental set-ups and two monochromatic light sources of different wavelengths are used to get fringes of equal width. The ratios of the slits separations and that of the wavelengths of light used are $2 : 1$ and $1 : 2$ respectively. The corresponding ratio of the distances between the slits and the respective screens ($D_1/D_2$) is ___________.
- The size of the images of an object, formed by a thin lens are equal when the object is placed at two different positions $8 \text{ cm}$ and $24 \text{ cm}$ from the lens. The focal length of the lens is ________ $\text{cm}$.
