Question

As shown in the figure, a combination of a thin plano concave lens and a thin plano convex lens is used to image an object placed at infinity The radius of curvature of both the lenses is $30 \,cm$ and refraction index of the material for both the lenses is $175$ Both the lenses are placed at distance of $40 \,cm$ from each other Due to the combination, the image of the object is formed at distance $x=$__ $cm$, from concave lens.

Answer 1

Correct Answer: 120

1. Focal length of the plano-concave lens (\(L_1\)): The focal length of a plano-concave lens is given by: \[ \frac{1}{f_1} = (1 - \mu) \left(\frac{1}{R}\right), \] where \(\mu = 1.75\) and \(R = 30 \, \text{cm}\). Substitute the values: \[ \frac{1}{f_1} = (1 - 1.75) \cdot \frac{1}{30} = -\frac{0.75}{30}. \] Thus: \[ f_1 = -40 \, \text{cm}. \]
2. Focal length of the plano-convex lens (\(L_2\)): The focal length of a plano-convex lens is given by: \[ \frac{1}{f_2} = (\mu - 1) \left(\frac{1}{R}\right). \] Substitute the values: \[ \frac{1}{f_2} = (1.75 - 1) \cdot \frac{1}{30} = \frac{0.75}{30}. \] Thus: \[ f_2 = 40 \, \text{cm}. \]
3. Final image formation: - The object is at infinity, so the image from \(L_1\) will be virtual and on the left at its focal length (\(f_1 = -40 \, \text{cm}\)). - This virtual image acts as the object for \(L_2\), which is placed \(40 \, \text{cm}\) away from \(L_1\). - The distance of the object from \(L_2\) is: \[ u = -40 - 40 = -80 \, \text{cm}. \] Using the lens formula for \(L_2\): \[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f_2}. \] Substitute the values: \[ \frac{1}{v} - \frac{1}{-80} = \frac{1}{40}. \] Simplify: \[ \frac{1}{v} + \frac{1}{80} = \frac{1}{40}. \] \[ \frac{1}{v} = \frac{1}{40} - \frac{1}{80} = \frac{2 - 1}{80} = \frac{1}{80}. \] Thus: \[ v = 80 \, \text{cm}. \] The final image is formed \(80 \, \text{cm}\) to the right of \(L_2\). From the concave lens \(L_1\), the total distance is: \[ x = 40 + 80 = 120 \, \text{cm}. \]