Question

An object of height 2 cm is held at a distance of 40 cm in front of a concave mirror of focal length
15 cm? Find the nature, position and size of the image?

Answer 1

$\huge\sf\pink{Answer}$

☞ Position of image is beyond the centre of curvature

☞ Nature of image is enlarged,and real

☞ Size is -1.2 cm

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$\huge\sf\blue{Given}$

✭ Height of object = 2 cm

✭ Distance of object = 40 cm

✭ Mirror = Concave

✭ Focal length = 15 cm

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$\huge\sf\gray{To \:Find}$

◈ Position of image?

◈ Size of image?

◈ Nature of image?

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$\huge\sf\purple{Steps}$

As the object is placed in front then the object distance & the focal length will be considered negative

So here to find the answer we shall use the lens formula, that is,

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

Substituting the given values,

➝ $\sf \dfrac{1}{v} + (-\dfrac{1}{40})= - \dfrac{1}{15}$

➝ $\sf \dfrac{1}{v} = - \dfrac{1}{15} + \dfrac{1}{40}$

➝ $\sf \dfrac{1}{v} = \dfrac{-40+15}{15\times 40}$

➝ $\sf \dfrac{1}{v} = \dfrac{-25}{600}$

➝ $\sf \dfrac{1}{v} = - \dfrac{1}{24}$

➝ $\sf \red{v = -24}$

We know that,

$\underline{\boxed{\sf\dfrac{I}{O} = -\dfrac{v}{u}}}$

Substituting the values,

➳ $\sf \dfrac{I}{2} = \dfrac{-(-24)}{-40}$

➳ $\sf \dfrac{I}{2} = \dfrac{24}{-40}$

➳ $\sf I = 2\times (-\dfrac{24}{40})$

➳ $\sf I = \dfrac{24}{20}$

➳ $\sf\orange{I = 1.2 \ cm}$

Nature of object will be diminished,real & inverted

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Answer 2

AnswEr:-

Your Answers Are:-

• Nature of image:- Real and Inverted.

• Position of image:- 24 cm in front of the mirror.

• Size of the image:- -1.2 cm. (as the image is inverted)

ExplanaTion:-

Given:-

• A concave mirror and an object.

• Height of a object (ho):- 2 cm.

• Object distance (u) = 40 cm.

• Focal length (f) = 15 cm.

To Find:-

• Nature of the Image.

• Position of the image (v).

• Height of the image (hi).

Concept Used:-

• Distance of the object from the concave mirror is always negative.

• Focal length of concave mirror is also always negative.

↦ u = -40 cm.

↦ f = -15 cm.

Formulas Used:-

1. Mirror Formula:-

$\blue{ \longrightarrow \boxed{ \red{ \bf \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}.}}}$

2. Magnification (m):-

$\blue{ \longrightarrow \boxed{ \red{ \bf m = -\dfrac{v}{u} = \frac{hi}{ho}.}}}$

Now,

By Using formulas (1) we get,

$\tt\mapsto \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} .$

$\tt\mapsto \dfrac{1}{v} + \dfrac{1}{ - 40 \: cm} = \dfrac{1}{ - 15 \: cm} .$

$\tt\mapsto \dfrac{1}{v} - \dfrac{1}{40 \: cm} = - \dfrac{1}{15 \: cm} .$

$\tt\mapsto \dfrac{1}{v} = - \dfrac{1}{15 \: cm} + \dfrac{1}{40 \: cm}.$

$\tt\mapsto \dfrac{1}{v} = \dfrac{ - 8 + 3}{120 \: cm} \: \: \rm(by \: \: taking \: \: lcm).$

$\tt\mapsto \dfrac{1}{v} = \dfrac{ \cancel{- 5 }\: \: \: {}^{ - 1} }{ \cancel{120 } \: cm\: \: \: _{24 \: cm}}.$

$\tt\mapsto \dfrac{1}{v} = - \dfrac{1}{24\: cm}$

$\large \red{ \mapsto \boxed{ \blue{ \bf v = - 24 \: cm.}}}$

Since the image distance is negative so it is in front of the mirror and 24 cm away from the mirror.

Similarly,

By Using Formula (2) we get:-

$\tt\mapsto m = -\dfrac{v}{u}.$

$\tt\mapsto m = -\dfrac{ \cancel{- 24} \: \: cm}{ \cancel{- 40 } \: \: cm}.$

$\tt\mapsto m = -\dfrac{3}{5}.$

$\large\red{ \mapsto \boxed{ \blue{ \bf m = -0.6}}}$

[Magnification is negative as the image is real and inverted]

Therefore The Magnification of the image is -0.6

Again,

By using formula (2) we get:-

$\tt\mapsto \dfrac{hi}{ho} = m.$

$\tt\mapsto \dfrac{hi}{2 \: cm} = -0.6$

$\tt\mapsto hi = -0.6 \times 2 \: cm.$

$\red{ \mapsto \boxed{ \blue{ \bf hi = -1.2 \: cm.}}}$

[Since The image is inverted so its height should be negative]

Therefore The Height Of The Image is 1.2 cm.

Therefore,

Nature of the image is real and inverted as it is in front of the mirror, Position of the image is 24 cm from the mirror and size of the image is 1.2 cm.