Question

A 5.0 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 20 cm. The distance of the object from the lens is 30 cm By calculation determine 

(i) The position and 

(ii) The size of the image formed.

Answer 1

According to the question;

Object distance = -30 cm; 

Image distance = v cm; 

Focal length = 20 cm 

By lens formula; 

\(\frac1v\,-\,\frac1u\,=\,\frac1f\)

⇒\(\frac1v\,-\,\frac{1}{-30}\,=\,\frac{1}{20}\) 

⇒\(\frac1v\,=\,\frac{1}{20}\,-\,\frac{1}{30}\)

⇒\(\frac1v\,=\,\frac{30-20}{600}\,-\,\frac{10}{600}\)

⇒\(\frac1v\,=\frac{1}{60}\)

v = 60cm.

Therefore image is formed at 60 cm in right of lens. 

Now; Height of object h₁= 5cm; 

Magnification = \(\frac{h_2}{h_1}\,=\,\frac{v}{u}\) 

Putting values of v and u 

Magnification = \(\frac{h_2}s =\,\frac{60}{-30}\)

⇒\(\frac{h_2}{s}\,=-2\) 

⇒ h₂ = - 2 × 5=10 

Height of image is 10 cm. 

Negative sign means image is real and inverted.