Question

An object of size 7.0 cm is placed at 27 CM in front of concave mirror of focal length 18 cm at what distance from the mirror should the screen be placed so that the short focus image can be obtained find the size and nature of the image

Answer 1

Focal length, f = – 18 cm

Object-distance, u = – 27 cm

Height of object, h1′= 5 cm

Image-distance, v = ?

Height of image h2′= ?

1 1 1

+

=

v u f

1 1 1

+

=

v − 27 -18 15

− 1 1 1

=

+

v 18 27

1 − 3 + 2 1

=

=

v 54 54

60

v =

= 54 cm

7

The screen should be kept at a distance of 54 cm in front of mirror

h2 − v

m =

=

h1 u

h2 (− 54 )

=> =

=

7 cm (− 27 )

(− 54 ) × 7 cm

=> h2 =

(− 27 )

Height (Size )

of Image

= h2 = 2 × 7 cm = 14 cm cm

Thus, a 14 cm high, virtual and an inverted image is formed

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.

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