# The image of an object place at 60 cm in front of a lens is obtained on a screen at a distance of 120 cm from it.

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The image of an object place at 60 cm in front of a lens is obtained on a screen at a distance of 120 cm from it. Find the focal length of the lens. What would be the lens? What would be the height of the image? If the object is 5 cm high?

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According to the question;

Object distance (u) = -60cm;

Image distance (v) = 120cm;

Focal length = f

By lens formula;

$\frac1v\,-\,\frac1u\,=\frac1f$

⇒ $\frac1{120}\,-\,\frac1{-60}\,=\frac1f$

⇒ $\frac1{f}\,=\,\frac1{120}\,+\frac1{60}$

⇒ $\frac1{f}\,=\,\frac{1+2}{120}\,+\frac3{120}$

⇒ f = $\frac{120}3$

⇒ f = 40cm.

Therefore Focal length of lens is 40 cm.

Since the focal length is positive. Therefore the lens is convex lens of focal length 40cm.

Now;

Height of object h1= 5cm;

Magnification =$\frac{h_2}{h_1}$ =$\frac{v}u$

Putting values of v and u

Magnification =$\frac{h_2}5$ = $\frac{120}{-60}$

$\frac{h_2}5$ = -2

⇒ h2 = -2 × 5= -10 cm

Height of image is 10 cm.

Negative sign means image is real and inverted.