An object is kept at a distance of 100 cm from each of the above lense

1.	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» (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

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»

(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