(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 1st case is -1. The negative sign means the image is real and inverted.
For 2nd 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