An object is placed at a distance of 10 cm from a convex mirror of focal length of 15 cm. Find the position and nature of the image.

Asked by Vishal kumar | 1 year ago |  353

Solution :-

Focal length of convex mirror (f) = +15 cm

Object distance (u) = - 10 cm

According to the mirror formula,

$$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$$

$$\frac{1}{v}=\frac{1}{15}-\frac{1}{-10} =\frac{2+3}{30}$$

$$v = \frac{5}{30}= 6 cm$$

$$Magnification =\frac{-v}{u}=\frac{-6}{-10}=0.6$$

The image is located at a distance of 6 cm from the mirror on the other side of the mirror. The positive and a value less than 1 of magnification indicates that the image formed is virtual and erect and diminished.

Answered by Shivani Kumari | 1 year ago

Related Questions

A 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?

A 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?

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

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.