Question

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?

Answer 1

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.