Question

An object of height 7 cm is kept at a distance of 25 cm in front of a concave mirror. The focal length of the mirror is 15 cm. At what distance from the mirror should a screen be kept so as to get a clear image? What will be the size and nature of the image

Answer 1

Object size (h₁) = 7 cm
Object distance (u) = -25 cm
Focal length (f) = -15cm
To find: Image distance (v) = ?
Image size (h₂) = ?
formula:
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}M=\frac{h_2}{h_1}=\frac{-v}{u}solution:\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{u}\Rightarrow \frac{1}{v}=\frac{1}{-15}-\frac{1}{-25}\Rightarrow \frac{1}{v}=\frac{1}{-15}+\frac{1}{25}\Rightarrow \frac{1}{v}=\frac{-5+3}{75}\Rightarrow v=\frac{-75}{2}\Rightarrow v=-37.5c.mThescreenshouldbekept37.5c.minfrontofthemirror.Theimageisreal.M=\frac{h_2}{h_1}=\frac{-v}{u}\Rightarrow h_2=\frac{-v\times h_1}{u}\Rightarrow h_2=\frac{-(-75)\times 7}{2\times -25}\Rightarrow h_2=\frac{-21}{2}\Rightarrow h_2=-10.5c.mTheheightoftheimageis10.5cm,itisaninvertedandenlargedimage.$