Question

A convex mirror has its radius of curvature 20 cm. Find the position of the image of a object placed at a distance of 12 cm from the mirror.​

Answer 1

From mirror formula,

u

1

+

v

1

=

f

1

;u is −ve (∵f=

2

R

=

2

20

=10cm)

v

1

=

f

1

−

u

1

=

10

−1

+

12

−1

=−(

60

5+6

)=

60

−11

(∵f is −ve for a concave mirror)

∴v=

11

60

cm

So the distance of the image is

6/11 and it is virtual ( -ve)

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Answer 2

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

• A convex mirror
• The radius of curvature is 20cm.
• The object placed at a distance of 1e cm from the mirror.

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

• The position of image

$\\ \large \: \bigstar \: \: \purple{ \underline{ { \underline{ \bf{Solution : - }}}} }$

Formula :-

$\\ \qquad \: \: \: \: \: \: \: \: \: \: \sf \: \underline{\boxed{ \sf \: \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }}$

Where,

• u is the object distance
• v is the image distance
• f is the focal length given by $\sf \: f = \dfrac{R}{2}$
• R is the radius of curvature of the convex mirror.

At first, Let's find the focal length

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \sf \: f = \dfrac{R}{2}$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \sf \: \rightarrow \: \dfrac{20}{2}$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \sf \: \bold{ f = 10cm}$

So, To find image distance the formula will be

$\\ \qquad \: \: \: \: \: \: \: \: \: \: \sf \: \sf \: \frac{1}{v} = \frac{1}{f} - \frac{1}{u}$

Now , put the given values in formula

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \implies \: \frac{ - 1}{10} + \frac{ - 1}{12}$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \implies \: - \frac{ 5}{60} + \frac{6}{60}$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \implies \: \bigg(\frac{ - 5 + 6}{60} \bigg)$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \implies \: \bigg(\frac{ - 11}{60} \bigg)$

Therefore,

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \implies \: \underline{\boxed{ \green{\frak{v = \frac{ 60}{11} }} } } \: cm$

$\\ \: \: \: \: \:† \: \sf So, \underline{\: the \: Image \: distance \: is \: \: \frac{60}{11} cm . }$