How far should an object be placed from a convex lens of focal length

1.	How far should an object be placed from a convex lens of focal length 20 cm to obtain its image at a distance of 30 cm from the lens? What will be the height of the image, if the objects is 6 cm tall?
Answer» According to the question; Object distance = u; Image distance (v) = 30cm; Focal length = 20cm By lens formula; \(\frac1v\,-\,\frac1u\,=\,\frac1f\) ⇒\(\frac{1}{30}\,-\,\frac1u\,=\,\frac{1}{20}\) ⇒\(\frac1u\,=\,\frac{1}{30}\,-\,\frac{1}{20}\) ⇒\(\frac1u\,=\,\frac{20\,-\,30}{600}\,=\,\frac{-10}{600}\) ⇒ \(\frac1u\,=\,\frac{-1}{60}\) u = -60cm. Therefore object is placed at 60 cm in front of lens. Now; Height of object h1= 6cm; Magnification =\(\frac{h_2}{h_1}\) =\(\frac{v}{u}\) Putting values of v and u Magnification =\(\frac{h_2}{6}\,=\,\frac{30}{-60}\) ⇒ \(\frac{h_2}{6}\,=\,\frac{-1}{2}\) ⇒ h₂ = - \(\frac62\) = -3 Height of image is 3 cm. Negative sign means image is real and inverted.

How far should an object be placed from a convex lens of focal length 20 cm to obtain its image at a distance of 30 cm from the lens? What will be the height of the image, if the objects is 6 cm tall?

Answer»

According to the question;

Object distance = u;

Image distance (v) = 30cm;

Focal length = 20cm

By lens formula;

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

⇒\(\frac{1}{30}\,-\,\frac1u\,=\,\frac{1}{20}\)

⇒\(\frac1u\,=\,\frac{1}{30}\,-\,\frac{1}{20}\)

⇒\(\frac1u\,=\,\frac{20\,-\,30}{600}\,=\,\frac{-10}{600}\)

⇒ \(\frac1u\,=\,\frac{-1}{60}\)

u = -60cm.

Therefore object is placed at 60 cm in front of lens.

Now;

Height of object h1= 6cm;

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

Putting values of v and u

Magnification =\(\frac{h_2}{6}\,=\,\frac{30}{-60}\)

⇒ \(\frac{h_2}{6}\,=\,\frac{-1}{2}\)

⇒ h₂ = - \(\frac62\) = -3

Height of image is 3 cm.

Negative sign means image is real and inverted.