Question

An object is kept at a distance of 100 cm from each of the above lenses. Calculate the

(a) image distance and 

(b) magnification in each of two cases.

Answer 1

(a) Given

Object distance (u) = -100cm 

Focal length (f) = 50cm

Image distance = v 

By lens formula;

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

⇒ \(\frac1v\,-\frac1{-100}=\frac1{50}\)

⇒\(\frac1v\,=\frac1{50}-\frac1{100}\)

⇒\(\frac1v\,=\frac{2-1}{100}-\frac1{100}\)

v =100cm 

Hence the image is formed at 100 cm behind the lens. 

Object distance (u) = -100cm 

Focal length (f) = -25cm 

Image distance = v 

By lens formula;

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

\(\frac1v\,-\frac1{-100}=\frac{-1}{25}\)

\(\frac1v\,-\frac{-1}{25}=\frac1{100}\)

\(\frac1v\,-\frac{-4-1}{-100}=\frac{-5}{100}\)

⇒ v = \(-\frac{100}5\)

⇒ v= -20cm 

Hence the image is formed at 20cm in front of the lens. 

(b) For 1st case 

Object distance (u) = -100cm 

Image distance (v) = 100cm. 

By Magnification Formula 

Magnification (m) =\(\frac{v}u\)

⇒ Magnification = \(\frac{100}{-100}\) 

Therefore, Magnification for the 1^st case is -1. The negative sign means the image is real and inverted. 

For 2^nd case 

Object distance (u) = -100cm 

Image distance (v) = -20cm. 

By Magnification Formula 

Magnification (m) = \(\frac{v}u\)

⇒ Magnification = \(\frac{-20}{-100}\,=\,\frac15\) = 0.5

Therefore, Magnification for 2nd case is 0.2. The positive sign means the image is erect and virtual