Height of the Object, h0 = 5 cm

Distance of the object from converging lens, u = -25 cm

Focal length of converging lens, f = 10 cm

Using lens formula,

\( \frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)

\( \frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{10}-\frac{1}{25}=\frac{15}{250}\)

\( v = \frac{250}{15}= 16.06 cm\)

Also, for a converging lens, \(\frac{h_i}{h_o}=\frac{v}{u}\)

\( h_i=\frac{v}{u}\times\,h_o=\frac{50\times5}{3\times(-25)}=\frac{10}{-3}= -3.3 cm\)

Thus, the image is inverted and formed at a distance of 16.7 cm behind the lens and measures 3.3 cm. The ray diagram is shown below.

Answered by Shivani Kumari | 2 years agoA doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?

Find the focal length of a lens of power -2.0 D. What type of lens is this?

An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed so that a sharply focused image can be obtained? Find the size and the nature of the image.

An object 5 cm is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position, nature, and size of the image.

The magnification produced by a plane mirror is +1. What does this mean?