Question

an object of size 7.0 is placed at 27cm in front of a concave mirror of focal length 18 cm . at what distance from measure should a screen be placed the so that a sharp focused image can be obtained? find the size of the nature of the image​

Answer 1

$\sf Given :$

• size of object, $\sf o=+7cm$
• distance of object , $\sf u=−27cm$
• Focal length of concave mirror, $\sf f= −18cm$
• let us take size of image = I

• so, mirror formula is

⟼ $\sf \frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

• so putting values of u and v

⟼ $\sf \frac{1}{v} = \frac{1}{ -18} - \frac{1}{ - 27}$

⟼ $\sf \frac{1}{v} = \frac{1}{ - 54}$

⟹ $\sf v = - 54cm$

• so image is formed on the object side only. magnification,

⟼ $\sf m = \frac{ - v}{u} = \frac{I}{o}$

⟼ $\sf m = - \frac{ - 54}{ - 27} = \frac{1}{7}$

⟹ $\sf I = - 14cm$

• so, Image is double in size to that of object.
• $\sf Nature:$ real, inverted and magnified image.

Answer 2

Answer:

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Explanation: