Question

Find the distance of the image when an object is placed on the principak axis at a distance of 5 cm in front of a concave mirror whose radius of curvature is 8cm?
please solve the problem step by step ​

Answer 1

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Given:

⚘ Radius of curvature = 8 cm

⚘ Focal length (f) = R/2

⠀⠀⠀⠀⠀⠀ ⠀⠀⠀= 8/2 = 4 cm

⚘ Object distance (u) = 5 cm

⚘ Image distance (v) = ?

Solution:

✾ Using Mirror Formula

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

$\frac{1}{4} = \frac{1}{5} + \frac{1}{v}$

$\frac{1}{v} = \frac{1}{4} - \frac{1}{5}$

$\frac{1}{v} = \frac{(5 + 4)}{20}$

$\frac{1}{v} = \frac{9}{20}$

$v = 20/9 cm$

$v = 2.22 cm$

❖ So the Image Distance will be $\boxed{2.22\:cm}$

Answer 2

Answer:-

v = 2.22 cm

Given :-

u = -5 cm

r = 8 cm

To find :-

The distance of the image.

Solution:-

As we know that,

The radius of curvature of a mirror is twice its focal length.

→$\huge \boxed{R = 2F}$

→$8 = 2f$

→$f =\mathsf{ \dfrac{8}{2}}$

→$\mathsf{f = 4 cm}$

Now,

The focal length of mirror is 4cm.

By using mirror formula :-

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

Put the given values,

→$\mathsf{\dfrac{1}{4} = \dfrac{1}{v}+\dfrac{-1}{5}}$

→$\mathsf{\dfrac{1}{4}=\dfrac{1}{v}-\dfrac{1}{5}}$

→$\mathsf{ \dfrac{1}{v}=\dfrac{1}{4}+\dfrac{1}{5}}$

→$\mathsf{ \dfrac{1}{v}=\dfrac{5+4}{20}}$

→$\mathsf{\dfrac{1}{v}=\dfrac{9}{20}}$

→$\mathsf{v = \dfrac{20}{9}}$

→$\mathsf{v = 2.22 cm}$

hence,

The image distance will be 2.22 cm.