Question

a concave mirror has a focal length of 10cm. at what distance from mirror should the object be placed so that it forms a real and inverted image 20 cm away from the mirror. what would be the size of image formed if the object is 2 cm high?​

Answer 1

Given :

➩ In Concave Mirror -

• Focal Lenght (f) = -10cm
• Image distance (v) = -20cm real and inverted image .
• Object size (ho) = 2cm

To Find :

• Object distance (u)
• Size of image (hi)

Formula Used :

$\bullet\underline{\boxed{\sf \dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}}}$

$\bullet\underline{\boxed{\sf Magnification \: (m)=\dfrac{h_i}{h_o}=\dfrac{-v}{u}}}$

hi = Height of image

ho = Height of object

Solution :

➞ Object Distance (u)

$\implies{\sf \dfrac{1}{-10}=\dfrac{1}{-20}+\dfrac{1}{u} }$

$\implies{\sf \dfrac{1}{u}=\dfrac{1}{-10}+\dfrac{1}{20} }$

$\implies{\sf \dfrac{1}{u}=\dfrac{20+(-10)}{-10 \times 20} }$

$\implies{\sf \dfrac{1}{u}=\dfrac{10}{-200} }$

$\implies{\sf u = \dfrac{-200}{10} }$

$\implies{\bf u = -20\: cm }$

Characteristics of image

• Image will be formed at centre of Curvature (C)
• Size of image is equal to the object size
• Image formed is real and inverted

➞ Size of image -

Size of image is equal to size of object i.e, 2cm because image is formed at centre of Curvature (C)

$\implies{\sf \dfrac{h_i}{h_o}=\dfrac{-v}{u}}$

$\implies{\sf \dfrac{h_i}{2}=\dfrac{-20}{20} }$

$\implies{\bf h_i = - 2\: cm}$

Answer :

Distance of Object (u) = -20cm

Size of image (hi) = -2cm