Question

An object 4 cm high is placed at a distance of 6 cm in front of a concave mirror of focal length 12cm.Find the position , nature and size of the image?

Answer 1

Given :

▪ Height of object (h) = 4cm

▪ Distance of object (u) = -6cm

▪ Focal length = -12cm

▪ Mirror type = Concave

To Find :

• Position of image
• Nature of image
• Size of image

Formula :

▶ Mirror formula :

$\underline{\boxed{\bf{\pink{\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}}}}}$

• X-coordinate of centre of curvature and focus of concave mirror are negative and those for convex mirror are positive.
• In case of mirrors since light rays reflect back in X-direction, therefore -ve sign of v indicates real image and +ve sign of v indicates virtual image.

▶ Lateral magnification :

$\underline{\boxed{\bf{\green{m=-\dfrac{v}{u}=\dfrac{h'}{h}}}}}$

• From the definition of m positive sign of m indicates erect image and negative sign indicates inverted image.

Calculation :

✴ Position of image :

✒ 1/v + 1/u = 1/f

✒ 1/v - 1/6 = -1/12

✒ 1/v = -1/12 + 1/6

✒ 1/v = -1 + 2/12

✒ v = 12cm

✴ Nature of image :

✏ m = -v/u

✏ m = -12/-6

✏ m = 2 (virtual and erect)

✴ Size of image :

→ m = h'/h

→ 2 = h'/4

→ h' = 2×4

→ h' = 8cm (enlarged)

Answer 2

GiveN :

• focal length = 12 cm.
• object distance = 6 cm.
• image distance = ?
• height of object = 4 cm.

SolutioN :

Let,

• f ( focal length )
• u ( object distance )
• v ( image distance )
• hi ( height of image )
• ho ( height of object )

★ Sign Convention.

• u = - 6 cm.
• f = - 12 cm.

✓ According to Mirror Formula.

$\tt \dagger \: \: \: \: \: \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$

» Putting given value.

$\tt : \implies\dfrac{1}{ - 12} = \dfrac{1}{v} + \dfrac{1}{ - 6}$

$\tt : \implies\dfrac{1}{ - 12} + \dfrac{1}{6} = \dfrac{1}{v}$

$\tt : \implies \dfrac{1}{v} = \dfrac{ - 1 + 2}{12}$

$\tt : \implies \dfrac{1}{v} = \dfrac{ 1 }{12}$

$\tt : \implies v = 12 \: cm.$

» Nature, Virtual and erect.

Now, Magnifications.

$\tt \dagger \: \: \: \: \: M = \dfrac{ - v}{u} = \dfrac{h_i}{h_o}$

$\tt : \implies \dfrac{ - 12}{ - 6} = \dfrac{ h_i}{4}$

$\tt : \implies \dfrac{ - 12 \times 4}{ - 6} = h_i$

$\tt : \implies h_i = 8 \: cm$

Therefore, the nature is virtual and erect and height of image is 8 cm.