Question

A candle which is 2.5 cm in height is placed 27 cm in front of a concave mirror having radius of curvature 36 cm. Find the distance from the mirror at which the screen should be placed to obtain a sharp image? Detail the size and nature of the image.

By how much the screen have to be moved if the candle is moved towards the mirror?

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Answer 1

SOLUTION :-

Height of the candle, h =2.5 cm

Image size=h’

Object distance, u= -27 cm

Radius of the concave mirror, R= -36 cm

Focal length of the concave mirror, f = R/2 = -36/2

= −18cm.

Image distance= v

=>1/v + 1/u = 1/f -------(Lens formula)

=>.°. 1/v= 1/f- 1/u

=>1/v = 1/-18 - 1/ -27

=>1/v= -3 + 2/54

=>1/v = -1/54.

=> .°. v = -54cm.

Therefore, to obtain a sharp image, the distance between the screen and the mirror should be 54cm.

Image magnification is:

m = h'/h = - v/u

h' = - v/u × h

h' = -(-54)/ -27 × 2.5

h' = -2 × 2.5

h' = -5cm.

5cm is the height of the image of the candle. As there is a negative sign, the image is inverted and virtual.

If the candle is moved closer to the mirror, then the screen needs to be moved away from the mirror so as to obtain the image.

Answer 2

★ Question ★

★ A candle which is 2.5 cm in height is placed 27 cm in front of a concave mirror having radius of curvature 36 cm. Find the distance from the mirror at which the screen should be placed to obtain a sharp image? Detail the size and nature of the image.

By how much the screen have to be moved if the candle is moved towards the mirror?

$\rule{300}2$

→height of candle =h1=2.5cm

→u=-27cm

→radius = -36cm

→focal length

$\implies \frac{r}{2} \\ \implies \frac{-36}{2} \\ \implies -18$

so,

$\implies\frac{1}{V}+\frac{1}{u}=\frac{1}{f} \\ \implies \frac{1}{v}=\frac{1}{f}-\frac{1}{u} \\ \implies \frac{1}{-18}-\frac{1}{-27} \\ \implies \frac{-1}{54} \\\implies v=-54cm$

so,

→screen should be placed 54cm from the mirror on the same side as the object.

$\rule{300}2$

*let us supposed

→h2 size of image .

so ,

magnification = $\implies\frac{h2 }{h1} =\frac{-v}{u} =\frac{v}{u} \\\implies\frac{-h2}{2.5}=\frac{-54}{-27} \\ \implies h2 = -5cm$

→there is a(-) negative sign i.e. image is inverted .

→ case less than <18cm = images virtual

so,

→virtual image cannot be placed on the screen.

$\rule{300}2$