Question

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Answer 1

➭ Question :-

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

➭ Given :-

⟶ object distance, u = -27cm

⟶ focal length, f = -18cm

⟶ Height of object, h = 7.0cm

➭ To find :-

⟶ Image distance, v

⟶ Height of image, h'

⟶ The size and nature of the image

➭ Solution :-

using mirror formula,

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

$\longmapsto \rm \frac{1}{v} + \frac{1}{ - 27} = \frac{1}{ - 18}$

$\longmapsto \rm \frac{1}{v} - \frac{1}{27} = - \frac{1}{18}$

$\longmapsto \rm \frac{1}{v} = - \frac{1}{18} + \frac{1}{27}$

$\longmapsto \rm \frac{1}{v} = \frac{ - 3 + 2}{54}$

$\longmapsto \rm \frac{1}{v} = - \frac{1}{54}$

$\longmapsto \boxed{ \rm v = - 54cm}$

The screen be placed at 54 cm in front of the concave mirror (on its left side)

→ The nature of image :-

• image will be obtained on the screen
• image is real
• image is inverted

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For the mirror,

$\longmapsto \rm m = \frac{h'}{h} = - \frac{v}{u}$

where,

• h = Height of object
• h' = Height of image
• v = image distance
• u = object distance

$\longmapsto \rm \frac{h'}{7.0} = - \cancel \frac{ - 54}{ - 27}$

$\longmapsto \rm \frac{h'}{7.0} = - 2$

$\longmapsto \rm h' = - 2 \times 7.0$

$\longmapsto \boxed {\rm h' = - 14.0cm}$

Hence, the size of image is 14cm