Question

A 5 CM tall object placed perpendicular to the principal axis of concave mirror of focal length 15 CM the distance of the object from the mirror is 20cm determines​

Answer 1

Correct Question:-

A 5 cm tall object is placed perpendicular to the principal axis of Concave mirror of focal length 15 cm. The distance of the object from the mirror is 20 cm. Find the position, nature and size of the image formed.

Given:-

• Height of the object is 5 cm.
• Focal length of the mirror is 15 cm.
• The distance between object and the mirror is 20 cm.

To Find:-

• Find the position, nature and size of the image formed ?

Assumption:-

Let, Object-distance be u.

       Image-distance be v.

       Focal-Length be f.

       Image-height be $\sf{h_e}$.

       Object-height be $\sf{h_o}$.

Solution:-

Now, At first we find position of the image by using certain formula

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

$\sf{Here,\ u = -20\ cm\ \ \ and\ \ f = -15\ cm.}$

Putting all the values we get,

$\sf{\Longrightarrow \dfrac{1}{v} + \dfrac{1}{-20} = \dfrac{1}{-15}}$

$\sf{\Longrightarrow \dfrac{1}{v} = \dfrac{1}{-15} - \dfrac{1}{-20}}$

$\sf{\Longrightarrow \dfrac{1}{v} = \dfrac{-4 + 3}{60}}$

$\sf{\Longrightarrow \dfrac{1}{v} = \dfrac{-1}{60}}$

By cross multiplication we get,

$\sf{\Longrightarrow -v = 60}$

$\sf{\Longrightarrow v = -60}$

So, the image is formed on the same side of the mirror and 60 cm apart from the mirror.

[Note:- Negative sign shows the image is formed in the same side of the object or mirror.]

Now, We find height of the image by using certain formula,

$\sf{\Longrightarrow \dfrac{h_e}{h_o} = \dfrac{-v}{u}}$

$\sf{Here,\ h_e = 5\ cm\ \ and\ \ v = -60\ cm\ \ and\ \ u = -20\ cm.}$

Putting all the values we get,

$\sf{\Longrightarrow \dfrac{h_e}{5} = \dfrac{-(-60)}{-20}}$

$\sf{\Longrightarrow \dfrac{h_e}{5} = \dfrac{60}{-20}}$

$\sf{\Longrightarrow \dfrac{h_e}{5} = -3}$

$\sf{\Longrightarrow h_e = -3 \times 5}$

$\sf{\Longrightarrow h_e = -15\ cm}$

Hence, Height of the image is 15 cm.

And, Nature of the image formed $\longrightarrow$ Real, Inverted and Enlarged.

[Note:- The Negative sign shows that the image is real and inverted ]

Answer 2

CORRECT QUESTION:-

A 5 cm tall object is placed perpendicular to the principal axis of Concave mirror of focal length 15 cm. The distance of the object from the mirror is 20 cm. Find the position, nature and size of the image formed.

FINAL ANSWER:-

• the position of image ($v$) ⇒ $-60cm$

• nature of image ⇒ real and inverted also image is larger than the object

• size of image ($hi$) ⇒ $-6cm$

GIVEN:-

• 5 CM tall object placed perpendicular to the principal axis of concave mirror

• focal length 15 CM

• distance of the object from the mirror is 20cm

TO FIND:-

• the position of image (V)

• nature of image

• size of image ($hi$)

THINGS TO ASSUME:-

distance of object ⇒ u

distance of image ⇒ v

focal length of mirror ⇒ $f$

height of object ⇒ $ho$

height of image ⇒ $hi$

SOLUTION:-

$ho=5cm\\f=-15cm\\u=-20cm\\\\using\:mirror \:formula\:we \\\:get\:position \:of\:image from\:mirror :-\\\\\frac{1}{f}= \frac{1}{v}+ \frac{1}{u}\\now \: put\:values\:in\:equation\\\frac{1}{-15}=\frac{1}{v}+\frac{1}{-20}\\\frac{1}{v}=\frac{-1}{15}+\frac{1}{20}\\\frac{1}{v}=\frac{-1*4+1*3}{60} \\\frac{1}{v}=\frac{-4+3}{60} \\\frac{1}{v}=\frac{-1}{60} \\v=-60cm\\\\put \:values\:in\:equation:-\\m=\frac{hi}{ho}=\frac{-v}{u} \\\frac{hi}{5}=\frac{-(-60)}{-50}\\hi=\frac{-60*5}{50}\\hi=-6cm$

size of image :-

$m=\frac{hi}{ho}\\m=\frac{-6}{5} \\m=-1.2$

• the position of image ($v$) ⇒ $-60cm$

• nature of image ⇒ real and inverted also image is larger than the object

• size of image ($hi$) ⇒ $-6cm$

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THINGS TO KNOW:-

• In a concave mirror, when the distance of the object is less than the focal length, the magnification will be greater than one. When the distance of the object is greater than the focal length, then the magnification is less than one.

• in concave mirror u and f both are always '-'ve. as they are on the left side of the mirror

• $ho$ (height of object) is always '+'ve and

• if $hi$ is '-'ve means the image is formed bottom of principal axis also image will be real and inverted and if $hi$ is '+'ve it means that image is virtual and erect.