Question

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

Answer 1

Answer:

Given :

Focal length of concave mirror (f)= -18cm

Conventionally concave mirror and concave lens have negative focal lengths and convex mirror and convex lens have positive focal lengths. The co-ordinate system is taken with respect to the Pole as origin. All distances are measured from the pole and distance measured in the direction of light rays are positive and opposite to the direction are considered as negative. Distances above the principal axis are positive and vice-versa. Following these conventions

Object distance (u)= -27cm

Object height= 7cm

Let image distance be v cm

We know by mirror formula that

1/v + 1/u = 1/f

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

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

1/v= (-3+2)/54

1/v= -1/54

v= -54 cm

Thus, placing a screen at a distance of 54 cm from the concave mirror on the same side as object will give us a sharp and focused image.

The magnification will be

m= -v/u

m= -(-54)/(-27)

m= -2

|m|>1 => enlarged image and the negative sign implies that the image is inverted

m can also be written as

m= Height of image/ Height of object

-2= Height of image/7

Height of image= -14cm

Thus the image has a height of 14 cm and is inverted and present at a distance of 54 cm from the mirror

Explanation:

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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