Question

An object of size 5 cm is placed at a distance of 25 cm from the pole of a concave mirror of radius of curvature 30 cm. Calculte the distance and size. Also tell the nature of the image

Answer 1

$\sf\star\bold\red{\underline{{QUESTION:-}}}$

✓✓An object of size 5 cm is placed at a distance of 25 cm from the pole of a concave mirror of radius of curvature 30 cm. Calculate the distance and size. Also tell the nature of the image.

$\sf\star\bold\green{\underline{{GIVEN:-}}}$

• $\sf\: Object\:size\:(h)\:=5cm$

• $\sf\: Object\: distance\:(u)\:=-25cm$

• $\sf\: Radius\:of\: curvature\:(R)\:=-30cm$ [R is negative for a concave mirror]

$\sf\therefore\:Focal\: Length\:=\dfrac{R}{2}$

$\sf\implies\:f=\dfrac{-30}{2}$

$\sf\implies\:f=-15cm$

$\sf\star\bold\orange{\underline{{SOLUTION:-}}}$

$\sf\boxed\star\pink{\underline{\underline{{{\bold\:Using\:the\:mirror\: formula:-}}}}}$

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

so, we have

$\sf\implies\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}$

$\sf\implies\dfrac{1}{v}=\dfrac{1}{-15}-\dfrac{1}{-25}$

$\sf\implies\dfrac{1}{v}=\dfrac{-5+3}{75}$

$\sf\implies\dfrac{1}{v}=\dfrac{-2}{75}$

$\sf\implies\:v=-37.5cm$

$\sf\pink{{magnification\:(m)}}=\dfrac{h'}{h}=-\dfrac{v}{u}$

$\sf\therefore\:Image\:Size\:,h'=-\dfrac{vh}{u}$

$\sf\implies\:h'=-\dfrac{(-37.5cm)(5.0cm)}{(-25cm)}$

$\sf\green{{\implies\:h'=-7.5cm}}$

$\sf\therefore\:$ As v is negative so, a real ,inverted image of height is 7.5cm is formed at a distance of 37.5cm in front of the mirror.