Explanation:
It is given that,
Height of the object, h = 3 mm = 0.3 m
Object distance, u = -16 cm
Radius of curvature, R = 16 cm
Focal length, f = R/2 = 8 cm
(b) The image distance is calculated using mirror's formula as:
[tex]\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}[/tex]
[tex]\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}[/tex]
[tex]\dfrac{1}{v}=\dfrac{1}{8}-\dfrac{1}{(-16)}[/tex]
v = 5.33 m
(c) The size of image is calculated using the formula of magnification as :
[tex]m=\dfrac{-v}{u}=\dfrac{h'}{h}[/tex]
h' is the size of image
[tex]h'=\dfrac{-vh}{u}[/tex]
[tex]h'=\dfrac{-5.33\times 0.3}{-16}[/tex]
h' = 0.3 m
Hence, this is the required solution.
