Question

An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18cm. At what distance from the mirror should be placed so that a sharp image can be obtained? Find the nature and size of the image?

Answer 1

A virtual image of the 14 cm size formed on the same side of mirror

We're having u = -27 , h = 7 and f = -18...and v = ?

Lets find out v

1/f = 1/u + 1/v

1/-18 = 1/-27 + 1/v

1/-18 + 1/27 = 1/v

-3+2/54 = 1/v

v = -54

Thus the image is formed at a distance of 54 cm in front of concave mirror(minus sign shows that the image is formed left side of the mirror)

The nature is real and inverted because it is formed in front of the concave mirror.

h1/h = -v/u

h1/7 = -54/-27

h1 = -14 cm

The size of the image is 14 cm and minus sign shows that it is formed infront of the concave mirror

mark as brainlist

Answer 2

•GIVEN:-

$\bf\:➠ Object \: distance \: ,u=-27cm$

$\bf {➠Object \: height,h=7cm}$

$\bf ➠Focal \: length=-18cm$

$\bf \large\underline{\overline{FORMULA \: USED⇓⇓}}$

$\bf \boxed{➣ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} }$

SOLUTION :-

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

$\frac{1}{v} = \bf \: \frac{1}{( - 18)} - \frac{1}{ (- 27)}$

$\frac{1}{v} = \bf \: \frac{1}{( - 18)} + \frac{1}{27}$

$\frac{1}{v} = \frac{( - 3 + 2)}{54} = \frac{ - 1 \: \: }{54 }$

$\boxed { \therefore \: v = \bf \: 54cm }$

•The screen should be placed at a distance of 54cm in front of the given mirror

$\bf \: ➽Magnification(m) = \frac{Image \: distance}{Object \: distance}$

$\bf \: m = \frac{ - 54}{ - 27} = - 2$

•Negative value of magnification shows that image is real.

$\bf \: m = \frac{Height \: of \: image}{Height \: of \: object} = \frac{h_1}{h}$

$\bf ➠h_1=m×h=(-2)×7=-14cm \\ \\ </p><p></p><p>$

➽The height of the image is negative which shows that image is inverted.

☆Drag your screen ➠ to view full answer.