Physics: An object 20mm high is placed 10cm from a sperical mirror and forms a virtual image which is 40mm high. what is the radius of curvature off the mirrors

An object 20mm high is placed 10cm from a sperical

dbanks701

09.08.2021

R = 40 cm

Explanation:

From the formulae of magnification:

$\frac{q}{p}=\frac{image\ height}{object\ height}\\\\\frac{q}{10\ cm} = \frac{4\ cm}{2\ cm}\\\\q = (10\ cm)(2)\\\\q = 20\ cm$

where,

q = image distance from mirror

p = object distance from mirror

Using thin lens formula:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}\\\\\frac{1}{f}=\frac{1}{10\ cm}+\frac{1}{-20\ cm}\\\\\frac{1}{f} = 0.05\\\\f = 20\ cm$

q is negative for the virtual image.

Now, the radius of the spherical mirror is double the focal length (f):

R = 2f

R = 2(20 cm)

R = 40 cm

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