Question

Find the distance of the image when an object is placed on the principal axis at a distance of 10 cm in front of a concave mirror of radius 8 cm

Answer 1

$\Large{\underline{\underline{\mathfrak{\bf{Question}}}}}$

Find the distance of the image when an object is placed on the principal axis at a distance of 10 cm in front of a concave mirror of radius 8 cm ?

$\Large{\underline{\underline{\mathfrak{\bf{Solution}}}}}$

$\Large{\underline{\mathfrak{\bf{Given}}}}$

• Distance of object from concave mirror (U) = -10 cm
• Radius of curvature of concave mirror ( R ) = -8 cm = 2F
• So, F = -8/2 = -4 cm

$\Large{\underline{\mathfrak{\bf{Find}}}}$

• Distance of image (V) .

$\Large{\underline{\underline{\mathfrak{\bf{Explanation}}}}}$

$\Large{\underline{\mathfrak{\bf{Using\:mirror\:formula}}}}$

$\red{\boxed{\sf{\orange{\:\dfrac{1}{F}\:=\:\dfrac{1}{U}\:+\:\dfrac{1}{V}}}}}$

Substitute value of U and F in above equation ,

$:\implies\sf{\:\dfrac{1}{V}\:+\:\dfrac{1}{-10}\:=\:\dfrac{1}{-4}} \\ \\ :\implies\sf{\:\dfrac{1}{V}\:=\:-\:\dfrac{1}{4}\:+\:\dfrac{1}{10}} \\ \\ :\implies\sf{\:\dfrac{1}{V}\:=\:\dfrac{-10+4}{40}} \\ \\ :\implies\sf{\:\dfrac{1}{V}\:=\:\dfrac{-14}{40}} \\ \\ :\implies\sf{\:V\:=\:\dfrac{-40}{14}} \\ \\ :\implies\sf{\red{\:V\:=\:\dfrac{-20}{7}\:cm}}$

$\Large{\underline{\mathfrak{\bf{Thus}}}}$

• Distance of image (V ) = -20/7 cm

When, the image distance is negative , the image is behind the mirror .

So, the image is virtual and upright.

Answer 2

Given

1. Distance from mirror = -10cm
2. radius=-8cm
3. f=-8/2=-4

TO FIND,

DISTANCE OF IMAGE=?

$\rule{300}2$

Solution→

$\frac{1}{v}+\frac{1}{-10}=\frac{1}{-4}\\\frac{1}{v}=\frac{-10+4}{40}\\\frac{1}{v}=\frac{-14}{40}\\v=\frac{-40}{14}\\v=\frac{-20}{7}$