Question

a bright object 50 mm high stands on the axis of a conave mirror of focal lenght 10pmm and a t a distance of 300mm frm the concave mirror . how big will the image be?

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Answer 1

Explanation:

using mirror formula

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

=> 1/ v - 1/10 = -1/300

=> 1/v = -1/300 + 1/10

=> 1/v = 29/300

=> v = 300/29

Now

$m = \frac{ - v}{u}$

=> m = - (300/29) ÷ 300

=> m = -1/29 (-ve sign means inverted image)

therefore size of image is 50× 1/29 = 50/29mm = 1.7241 mm