A bi-convex lens is formed with two thin plan | vedclass

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(B) The focal length $f$ of a lens is given by the lens maker's formula: $\frac{1}{f} = (n-1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$.For the fi

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$40.0$

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(B) The focal length $f$ of a lens is given by the lens maker's formula: $\frac{1}{f} = (n-1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$.For the first plano-convex lens $(n_1 = 1.5)$: $\frac{1}{f_1} = (1.5 - 1) \left[ \frac{1}{14} - \frac{1}{\infty} \right] = \frac{0.5}{14} = \frac{1}{28} \ cm^{-1}$.For the second plano-convex lens $(n_2 = 1.2)$: $\frac{1}{f_2} = (1.2 - 1) \left[ \frac{1}{\infty} - \frac{1}{-14} \right] = \frac{0.2}{14} = \frac{1}{70} \ cm^{-1}$.For a combination of thin lenses in contact,the effective focal length $F$ is given by $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$.$\frac{1}{F} = \frac{1}{28} + \frac{1}{70} = \frac{5 + 2}{140} = \frac{7}{140} = \frac{1}{20} \ cm^{-1}$.Thus,the effective focal length $F = 20 \ cm$.Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{F}$,where $u = -40 \ cm$:$\frac{1}{v} - \frac{1}{-40} = \frac{1}{20} \implies \frac{1}{v} = \frac{1}{20} - \frac{1}{40} = \frac{2 - 1}{40} = \frac{1}{40}$.Therefore,the image distance $v = 40 \ cm$.

Similar Questions

$A$ convex lens is in contact with a concave lens. The magnitude of the ratio of their focal lengths is $2/3$. Their equivalent focal length is $30 \, cm$. Their individual focal lengths in $cm$ will be

$A$ convex lens of focal length $40 \, cm$ and a concave lens of focal length $25 \, cm$ are placed in contact. The power of the combination is .......... $D$.

The position of the final image formed by the given lens combination from the third lens will be at a distance of $(f_1 = +10 \ cm, f_2 = -10 \ cm, f_3 = +30 \ cm)$.

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