Question

If we want an inverted image of height 10 cm of an object kept at a distance of 50cm from a concave mirror. The focal length of the mirror is 20cm. At what distance from the mirror should we place the screen? What will be the type of the image, and what is the height of the object ?

Answer 1

Answer:

Distance of mirror from the screen = -100 / 3 cm = -33.3 cm

Height of object = 15 cm

Nature of image = real , inverted & diminished.

Explanation:

Height of image = -10 cm (as image is real)

Focal length (f) = -20 cm

Object distance (u) = -50 cm

From mirror formula ,

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

⇒ $-\frac{1}{20} = \frac{1}{v} - \frac{1}{50}$

⇒ $\frac{1}{v} = \frac{1}{50} - \frac{1}{20}$

⇒$\frac{1}{v} = \frac{2-5}{100}$

⇒$\frac{1}{v} = \frac{-3}{100}$

⇒v = Image distance = $-\frac{100}{3}$ cm = -33.3 cm

Magnification = $-\frac{v}{u}$ = -($\frac{100}{150}$) = -2/3

$-\frac{v}{u}$ = $H_{i} /H_{o}$

⇒ $-\frac{10}{H_o} = -\frac{2}{3}$

⇒$H_o$ = $\frac{3}{2}$ × 10 = 15 cm

Negative value of magnification & image distance indicates that the image is real & inverted.

As M < 1 the image is diminished.

∴ The nature of image is real , inverted & diminished.