Question

an object 5cm high is placed20cm in front of a concave mirror of focsl length 1.5cm.At distance from a mirror should a screen be placed to obtain a sharp image?calculate the size of the image formed​

Answer 1

GIVEN :

• Height of the object (h1) = 5 cm.
• Height of the image (h2) = ?
• Focal length of mirror (f) = -15 cm.
• Object distance (u) = -20 cm.
• Image distance (v) = ?

FORMULA :

• $\sf \dfrac {1}{f} \: = \: \dfrac {1}{v} \: + \: \dfrac {1}{u}$
• Magnification (m) = $\sf \dfrac {height \: of \: image}{height \: of \: object}$
• Magnification (m) = $\sf \dfrac {-v}{u}$

TO FIND :

• Image distance (v) = ?
• Height of the image (h2) = ?

SOLUTION :

Using formula to find image distance,

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

$\implies \sf \dfrac {-1}{15} \: = \: \dfrac {1}{v} \: - \: \dfrac {1}{20}$

$\implies \sf \dfrac {1}{20} \: - \: \dfrac {1}{15} \: = \: \dfrac {1}{v}$

$\implies \sf \dfrac {1}{v} \: = \: \dfrac {1}{60}$

$\implies {\boxed{\sf v \: = \: 60}}$

$\therefore$ The image will be formed 60 cm in front of the mirror.

Now, To find magnification (m),

We know that,

$\sf m \: = \: \dfrac {-v}{u}$

$\implies m \: = \: - \: \dfrac {-60}{-20}$

$\implies \sf m\: = \: 3$

Here,

$\sf m \: = \: \dfrac {h_2}{h1}$

$\implies \sf -3 \: = \: \dfrac {h_2}{5}$

$\implies \sf -15 \: = \: h_2$

$\implies {\boxed{\sf h_2 \: = \: -15}}$

$\therefore$ The size of the image is 12 cm in height and size is in negative sign, so image will formed below the principal axis.