A lens forms a real image 3cm high of an object 1cm high. If the separation of object and image is 15cm, find the focal length of the lens.
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Given, Height of image (h') = -3cm (negative because image is real) Height of object = 1cm Let us first define the sign convention, Since the object distance is always negative, let it be -u. Since the focal length is positive for a convex lens, let it be +f.$\text{Magnification}=\frac{\text{h}'}{\text{h}}=\frac{\text{v}}{\text{u}}$
$\frac{\text{v}}{\text{u}}=-3$
$\therefore\ \text{v}=-3\text{u}$
Since object distance (u) is always negative, and the ratio $\frac{\text{v}}{\text{u}}$ here is negative, v must be positive i.e., it is on the right side of the lens. Now, we have object distance + image distance = 15 v + (-u) = 15 (Using sign convention)$-3\text{u}-\text{u}=15$
$-4\text{u}=15$
$\text{u}=\frac{-15}{4}$
Applying the lens formula,$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$
$\frac{1}{\text{f}}=\frac{1}{-3\text{u}}-\frac{1}{\text{u}}$
$\frac{1}{\text{f}}=\frac{-4}{3\text{u}}$
$\text{f}=\frac{3\text{u}}{-4}$
$\text{f}=\frac{3\times\frac{-15}{4}}{4}=\frac{45}{16}$
$\text{f}=2.18\text{cm}$
Hence, the focal length of the lens is +2.18cm.
$\frac{\text{v}}{\text{u}}=-3$
$\therefore\ \text{v}=-3\text{u}$
Since object distance (u) is always negative, and the ratio $\frac{\text{v}}{\text{u}}$ here is negative, v must be positive i.e., it is on the right side of the lens. Now, we have object distance + image distance = 15 v + (-u) = 15 (Using sign convention)$-3\text{u}-\text{u}=15$
$-4\text{u}=15$
$\text{u}=\frac{-15}{4}$
Applying the lens formula,$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$
$\frac{1}{\text{f}}=\frac{1}{-3\text{u}}-\frac{1}{\text{u}}$
$\frac{1}{\text{f}}=\frac{-4}{3\text{u}}$
$\text{f}=\frac{3\text{u}}{-4}$
$\text{f}=\frac{3\times\frac{-15}{4}}{4}=\frac{45}{16}$
$\text{f}=2.18\text{cm}$
Hence, the focal length of the lens is +2.18cm.
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