Question

An object is placed 15cm from q convex mirror of radius of curvature 90cm calculate the positions and magnification ​

Answer 1

Answer⤵️

$r = 90cm \\ u = 15cm \\ f = - 45cm$

$v = \frac{u \times f}{u - f} \\ = \frac{15 \times ( - 45)}{15 - ( - 45)} \\ = \frac{ - 675}{15 + 45} \\ = \frac{ - 675}{60} \\ = - 11.25cm$

Image is imaginary at a distance of 11.25 cm from optical center of mirror.

• magnification

$\frac{ - v}{u} = \frac{11.25}{15} \\ = 0.75$

Coordinate System:

• formula:

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

• Distance measured in side where reflected rays exists in positive, elsewhere negative. (for all measurement except u)
• Moving from Object to Optical center, if direction is same as the direction of incident ray u is positive or else negative. (only applicable for u)

hope this may helps

Brainliest plz!