Answer to Question #269227 in Electricity and Magnetism for NICKO

Question

 An object 0.600 cm tall is placed 16.5 cm to the left of the vertex of a concave spherical mirror having a radius of curvature of 22.0 cm. (a) Draw a principal-ray diagram showing the formation of the image. (b) Determine the position, size, orientation, and nature (real or virtual) of the image.

Answer 1

The size of the object is 0.600 cm.

The distance of the object is 16.5 cm.

The radius of curvature is 22.0 cm.

(a)

(b)

Write the expression for the mirror formula.

$\frac{1}{v}+ \frac{1}{u}= \frac{2}{R}$

Substitute the values.

$\frac{1}{v}+\frac{1}{16.5} = \frac{2}{22.0} \\ v = \frac{22 \times 16.5}{2 \times 16.5 -22} \\ v = 33 \;cm$

Write the expression for the magnification.

$m=\frac{-v}{u}$

Substitute the values.

$m= \frac{-33}{16.5} = -2$

Calculate the height of the image.

$m=\frac{h'}{h}$

Substitute the values.

$-2=\frac{h'}{0.600} = -1.20 \; cm$

Thus, the position of the image is 33.0 cm from the mirror, the size of the image is 1.20 cm, the image is real and inverted.