Question

12.) If an object of 10 cm height is placed at a distance of 36 cm from a concave mirror of focal length 12 cm,find the position,nature and height of the image.​

Answer 1

$\large\bf\underline {To \: find:-}$

• we need to find the position, nature and height of image.

$\large\bf\underline{Given:-}$

• If an object of 10 cm height is placed at a distance of 36 cm from a concave mirror of focal length 12 cm

$\huge\bf\underline{Solution:-}$

• size of object (h) = 10cm
• object distance (u) = - 36cm
• Focal length (f) = -12cm
• image height (h') = ?

By using mirror Formula :-

$\: \: \: \: \: \: \huge{\dag} \large{ \underline{ \boxed{ \red{ \bf \: \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }}}}$

$\dashrightarrow \rm\: \frac{1}{v} = \frac{1}{f} - \frac{1}{u}$

$\dashrightarrow \rm\: \frac{1}{v} = \frac{1}{( - 12)} - \frac{1}{ ( - 36)}$

$\dashrightarrow \rm\: \frac{1}{v} = - \frac{ 1}{12} + \frac{1}{36}$

$\dashrightarrow \rm\: \frac{1}{v} = \frac{ - 3 + 1 \: }{36}$

$\dashrightarrow \rm\: \frac{1}{v} = \frac{ - 2}{36}$

$\dashrightarrow \rm\: \frac{1}{v} = \frac{ - 1}{18}$

$\dashrightarrow \rm\: - v = 18$

$\dashrightarrow \bf\:v = - 18$

we know that,

$\bf \star \green{m = \frac{h'}{h} = - \frac{ v}{u} }$

$\rightarrowtail \: \rm \frac{h'}{10} = - \frac{( - 18)}{ (- 36)}$

$\rightarrowtail \: \rm \frac{h'}{10} = - \frac{18}{36}$

$\rightarrowtail \: \rm \frac{h'}{10} = - \frac{1}{2}$

$\rightarrowtail \: \rm \: 2h' = - 10$

$\rightarrowtail \: \rm \: h' = \frac{ - 10}{2}$

$\rightarrowtail \: \bf \: h' = - 5cm$

Hence,

★The position of image is 18cm in front of mirror.

★ Height of image is -5cm

★The nature of the image is real and inverted .

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Answer 2

1/v + 1/u = 1/f

u = -36cm f = -12 cm

1/v = -1/12 + 1/36 = -1/18 so v= -18 cm

m = - v/u = -1/2

height of the image = 10 cm * -1/2 = -5 cm

so image is real inverted and of height 5 cm.