Question

a 1.4cm long object is placed perpendicular to the principal axis of a convex mirror of focal length 15cm at a distance of 10cm from it calculate the location of the image,
height of the image ,
nature of the image​

Answer 1

Answer:-

• Location Of Image = 6 cm behind the mirror.

• Height of the image = 0.84 cm.

• Nature of image = Virtual And erect.

Explanation:-

Given:-

• Height of the object (ho) = 1.4 cm

• Focal length of the convex mirror (f) = 15 cm.

• Object distance (u) = 10 cm.

To Find:-

• Location of the image (v).

• Height of the image (hi).

• Nature of the image.

Formulas Used:-

Mirror Formula:-

$\orange{1. \: \: \boxed{ \pink{ \bf \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} .}}}$

Magnification Formula:-

$\orange{2. \: \: \boxed{ \pink{ \bf m = - \dfrac{v}{u} = \dfrac{hi}{ho}. }}}$

So Here,

• v = ?? [To Find]

• u = -10 cm (As the object distance is always taken as negative).

• f = 15 cm (As the focal length of convex mirror is always positive).

• hi = ?? [To Find].

• ho = 1.4 cm

Now,

By putting the values in formula (1) we get:-

$\tt\mapsto \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} . \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \tt\mapsto \dfrac{1}{v} = \frac{1}{f} - \frac{1}{u} . \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \tt\mapsto \dfrac{1}{v} = \bigg( \frac{1}{15 \: cm} \bigg) - \bigg( \frac{1}{ - 10 \:cm}\bigg). \\ \\ \tt\mapsto \dfrac{1}{v} = \bigg(\frac{1}{15 \: cm} + \frac{1}{10 \: cm} \bigg) \: \: cm. \: \: \: \: \: \\ \\ \tt\mapsto \dfrac{1}{v} = \bigg( \frac{(1 \times 2) + (1 \times 3)}{30 \: cm} \bigg). \: \: \: \: \: \: \\ \\ \bf(by \: \: taking \: \: lcm). \\ \\ \tt\mapsto \dfrac{1}{v} = \bigg( \frac{2 + 3}{30 \: cm} \bigg). \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \tt\mapsto \dfrac{1}{v} = \cancel \frac{5}{30 \: cm}. \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \tt\mapsto \dfrac{1}{v} = \frac{1}{6 \: cm}. \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \large \blue{\mapsto \boxed{ \red{ \bf v = 6 \: cm.}}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Therefore Image Distance Is 6 cm and since it is positive so the image is formed behind the mirror.

Similarly,

By Putting The Values in formula (2) we get:-

$\tt\mapsto m = - \dfrac{v}{u} = \dfrac{hi}{ho}. \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \tt\mapsto - \frac{v}{u} = \frac{hi}{ho}. \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \tt\mapsto \frac{ \cancel{ -} \: \: 6 \: cm}{ \cancel{ - } \: \: 10 \: cm} = \frac{hi}{1.4 \: cm}. \\ \\ \tt\mapsto \cancel \frac{6 \: cm}{ 10 \:cm} = \frac{hi}{1.4 \: \: cm} . \: \: \: \: \: \\ \\ \tt\mapsto \frac{3 \times 1.4 \: cm}{5} = hi. \: \: \: \: \: \: \\ \\ \tt\mapsto hi = \frac{4.2 \: cm}{5} . \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \large \red{\mapsto \boxed{ \blue{ \bf hi = 0.84 \: cm}}} \: \: \:$

Therefore The Heigth Of The Image Is 0.84 cm.

Therefore,

• Image Position = 6 cm behind the mirror.

• Height of the Image = 0.84 cm.

• Nature of the Image = Virtual And Erect.