Question

An object is placed 20 cm from (a) a converging lens, and (b) a diverging lens, of focal length 15 cm. Calculate the image position and magnification in each case. 

Answer 1

) Focal length of the converging lens, i.e., convex lens f = + 15 cm
Object distance u = - 20 cm
Using the lens formula: 

`1/f=1/v-1/u` 

⇒`1/15=1/v-1/-20` 

⇒`1/15=1/v+1/20` 

⇒`1/v=1/15-1/20` 

⇒`1/v=(4-3)/60=1/60`  

∴ v = +60 cm.

Therefore, the image formed is real. It is at a distance of 60 cm from the lens and to its right. 

`"Magnification" ="image distance"/ "object distance"` 

∴ `m=v/u=60/-20=-3` 

The magnification is greater than 1. Therefore, the image is magnified.  
The negative sign shows that the image is inverted. 

(b) ​Focal length of the diverging lens, i.e., concave lens f = - 15 cm
Object distance u = - 20 cm

Using the lens formula: 

`1/f=1/v-1/u` 

⇒`1/-15=1/v-1/-20` 

⇒`1/v=1/-15-1/20` 

⇒`1/v=(-4-3)/60=-7/60` 

∴ v = - 8.57 cm.

Therefore, the image formed is virtual. It is at a distance of 8.57 cm from the lens and to its left. 

`"Magnification"="image distance"/"object distance"`  

∴ `m=v/u=(-8.57)/-20=+0.42` 

The magnification is less than 1. Therefore, the image is diminished. The positive sign indicates that the image is erect.