Question

An object 2 cm in size is placed 30 cm in front of a concave mirror of focal length 15 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? What will be the nature and the size of the image formed? Draw a ray diagram to show the formation of the image in this case.

Answer 1

Dear Student,

u = -30 cm , f = -15.0 cm , h0 = 2
$$\begin{gathered} U\sin fMirrorFormula, \\ \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \\ \frac{1}{v}-\frac{1}{30}=-\frac{1}{15} \\ \frac{1}{v}=-\frac{1}{15}+\frac{1}{30} \\ v=-30cm \\ m=\frac{hi}{ho}=-\frac{v}{u} \\ \frac{hi}{2}=-1 \\ hi=-2cm \end{gathered}$$ cm

Screen should be placed at a distance of 30 cm on the left side from the pole in order to obtain a sharp image.

Size of the image: 2 cm.

Nature of the image: Real and inverted.