Question

You are given a concave mirror of focal length 10 cm. An object in front of the concave mirror at a

distance equal to twice the focal length of the mirror. Where will the image be formed? Also state the

properties of image.​

Answer 1

Given:-

• Focal length ,f = 10cm

• Object Distance is twice of focal length = 2×10 = 20
• Object distance ,u = -20 cm

To Find:-

• Image Position ,v

Solution:-

Using mirror formula

• 1/v + 1/u = 1/f

where,

v is the Image Position

u is the object distance

f is the focal length

Substitute the value we get

→ 1/v = 1/f -1/u

→ 1/v = 1/10 -1/(-20)

→ 1/v = 1/10 +1/20

→ 1/v = 2+1/20

→ 1/v = 3/20

→ v = 20/3

→ v = 6.66 cm

Therefore, the image position is 6 cm .

The image position is between Focus & Center of Curvature.

Nature of image is Real & smaller than the object ,inverted .

Answer 2

${\large{\underline{\boxed{\bf{\pink{Given : }}}}}}$

• Concave mirror
• Focal Length = 10 cm
• An object in front of the concave mirror at a distance equal to twice the focal length of the mirror.

${\large{\underline{\boxed{\bf{\pink{To \: find : }}}}}}$

• Where will the image be formed?
• Properties of image.

${\Large{\underline{\boxed{\bf{\red{Solution : }}}}}}$

Concept of the question:-

• Here we have a question of Light - Reflection and Refraction in which we have given that A concave mirror of focal length of 10 cm. Object is in front of the mirror and distance is double of the focal length. We need to find where will object formed and properties of image.

So, we have :-

• Focal Length = 10 cm
• Object Distance = Double of Focal length = 10 × 2 = 20 cm
• Object Diatance = (-20) [-20 Because object is always placed in front of the mirror hence the sign of object is taken as negative.]

We need to find :-

• Image distance = ?

So, formula for finding Image distance is :-

${ \underline{\boxed{\Large{\bf{ \implies\frac{1}{v} + \frac{1}{u} = \frac{1}{f} }}}}}$

Here :-

• v = image distance
• u = Object distance
• f = focal length

■ So Image will formed at :-

$\\ {\bf{ \implies\frac{1}{v} + \frac{1}{u} = \frac{1}{f} }}$

$\\ {\bf{ \implies\frac{1}{v} + \frac{1}{ - 20} = \frac{1}{10} }}$

$\\ {\bf{ \implies\frac{1}{v} = \frac{1}{10} - \frac{1}{ - 20} }}$

$\\ {\bf{ \implies\frac{1}{v} = \frac{1}{10} + \frac{1}{ 20} }}$

$\\ {\bf{ \implies\frac{1}{v} = \frac{1 \times 2}{10 \times 2} + \frac{1}{ 20} }}$

$\\ {\bf{ \implies\frac{1}{v} = \frac{2}{20} + \frac{1}{ 20} }}$

$\\ {\bf{ \implies\frac{1}{v} = \frac{2 + 1}{20} }}$

$\\ {\bf{ \implies\frac{1}{v} = \frac{3}{20} }}$

$\\ {\bf{ \implies{1 \times 20 = 3 \times v }}}$

$\\ {\bf{ \implies{20 = 3 v }}}$

$\\ {\bf{\implies{v = \frac{20}{3} }}}$

${ \underline{\boxed{\Large{\red{ \bf{\implies{v = 6.67cm}}}}}}}$

● The image will formed at 6.67 cm at back of the mirror.

■ Properties of image formed by this concave mirror.

• The image formed is smaller than the object , real and inverted.

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