Question

An object is placed at a distance of 15 cm in front of a convex mirror of radius of curvature of 10 cm. Find :- (a) position of the image (b) the magnification (c) What will be the nature of the image real or virtual ?

Answer 1

Answer:

V = 15/4cm (position of image)

M = - v/u

= - 15/4

15

M = - 4 (magnification)

Virtual image will be formed.

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Answer 2

Answer:

Hey Zeshawn!!

Explanation:

So here is your answer-

u = (-)15

v = (+)?

f = (+)10/2 = (+)5

(a) Mirror formula =>

$\frac{1}{f} = \frac{1}{u} +\frac{1}{v}\\\\\frac{1}{5} = \frac{1}{-15} +\frac{1}{v}\\\\\frac{1}{5} - \frac{1}{-15} +\frac{1}{v}\\\\\frac{3}{15} - \frac{-1}{15}= \frac{1}{v}\\\\\frac{3-(-1)}{15}= \frac{1}{v}\\\\\ \frac{4}{15}= \frac{1}{v}\\\\\frac{15}{4}=v\\\\3.75 cm = v = position~of~the~image$

(b) Magnification = -v/u

$m = \frac{-v}{u}\\\\m = \frac{-3.75}{-15}\\\\m = 0.25$

(c) The nature will be Virtual image

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