Question

consider the situation where:
• an object is 3 cm
• mirror is concave with 6 focal length.
• object is placed at the centre of curvature.
which of the following options are correct?
A. the mirror will produce an image of magnification +1.5.
B. The mirror will produce an image of magnification -1.
C. The mirror will produce an image of magnification + 1.
D. The imarah will produce an image of magnification -1.5.​

Answer 1

• The object height is 3 cm.
• The object had been placed at the centre of curvature of a concave mirror of focal length 6 cm.

Now, let's find the image distance first:

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

$\implies \dfrac{1}{ - 6} = \dfrac{1}{v} + \dfrac{1}{ - 12}$

$\implies \dfrac{1}{v} = - \dfrac{1}{6} + \dfrac{1}{12}$

$\implies \dfrac{1}{v} = - \dfrac{1}{12}$

$\implies v = - 12 \: cm$

• So, image distance is -12 cm. That means image is produced at the centre of curvature.

$mag. = - \dfrac{v}{u} = - \dfrac{ - 12}{ - 12} = - 1$

So, magnification is -1.