Question

The focal length of a concave mirror is 10cm . Where should the object be placed in front of the mirror to obtain an image of half size of the object . Answer : 10 cm

Answer 1

GIVEN

◙ Focal length, f = -10 cm

◙ Image size is half of the object size.

⇒ h$\sf{_o}$ = 2h$\sf{_i}$

$\underline{\rule{190}4}$

TO FIND

The object distance (u) so that, the image size is half of the object size.

$\underline{\rule{190}4}$

WE MUST KNOW

◙ Magnification :-

$\displaystyle{\sf{m=\frac{-v}{u}=\frac{h_i}{h_o} }}$

◙ MIRROR FORMULA :-

$\boxed{ \displaystyle{\sf{\frac{1}{v}+\frac{1}{u}=\frac{1}{f} }}}$

$\underline{\rule{190}4}$

SOLUTION

Using the magnification equation for a concave mirror :-

$\displaystyle{\sf{\frac{-v}{u} =\frac{h_i}{h_o} }}\\\\\displaystyle{\sf{\dashrightarrow \frac{-v}{u}=\frac{h_i}{2h_i} }}\\\\\displaystyle{\sf{\dashrightarrow \frac{-v}{u}=\frac{1}{2} }}\\\\\displaystyle{\sf{\dashrightarrow -2v=u}}\:\:\:\: \cdots \sf{(i)}$

$\rule{140}2$

Now, using the mirror formula :-

$\displaystyle{\sf{\frac{1}{v}+\frac{1}{u}=\frac{1}{f} }}\\\\\displaystyle{\sf{\dashrightarrow }\frac{1}{v}+\frac{1}{-2v}=\frac{1}{-10} }\\\\\displaystyle{\sf{\dashrightarrow \frac{-2+1}{-2v} =\frac{1}{-10} }}\\\\\displaystyle{\sf{\dashrightarrow \frac{-1}{2v}=\frac{1}{-10} }}\\\\\displaystyle{\sf{\dashrightarrow 2v=10}}\\\\\displaystyle{\sf{\dashrightarrow v=\frac{10}{2} }}$

$\displaystyle{\sf{\dashrightarrow v=5\ cm}}$

Now, putting the value of v in eq. (i) :-

$\displaystyle{\sf{\dashrightarrow -2v=u }}\\\\\displaystyle{\sf{\dashrightarrow -2\times 5=u}}\\\\\large\boxed{\displaystyle{\sf{\dashrightarrow u=-10\ cm}}}\:\:\:\: \cdots \mathbf{ANSWER}$

Therefore, the object must be kept 10 cm in front of the concave mirror to obtain an image whose size is half the object size.

Answer 2

Given :

▪ Focal length = 10cm

▪ Height of image = Height of object ÷ 2

▪ Type of mirror : Concave

To Find :

▪ Distance of object (u)

Concept :

❇ X-coordinate of centre of curvature and focus of concave mirror are negative and those for convex mirror are positive.

❇ Lateral Magnification :

$\bigstar\:\underline{\boxed{\bf{\red{m=-\dfrac{v}{u}=\dfrac{h'}{h}}}}}$

❇ Mirror Formula :

$\bigstar\: \underline{\boxed{\bf{\blue{\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}}}}}$

• u denotes distance of object
• v denotes distance of image
• f denotes focal length
• h' denotes height of image
• h denotes height of object

Calculation :

$\implies\sf\:-\dfrac{v}{u}=\dfrac{(-h')}{h}\\ \\ \implies\sf\:-\dfrac{(-v)}{(-u)}=-\dfrac{1}{2}\\ \\ \implies\underline{\underline{\bf{v=\dfrac{u}{2}}}}\\ \\ \mapsto\sf\:\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}\\ \\ \mapsto\sf\:\dfrac{1}{(-u)}+\dfrac{2}{u}=\dfrac{1}{(-10)}\\ \\ \mapsto\sf\:\dfrac{1}{u}=-\dfrac{1}{10}\\ \\ \mapsto\underline{\boxed{\bf{\green{u=-10cm}}}}\:\orange{\bigstar}$