A 5 CM tall object placed perpendicular to the principal axis of conca

1.	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» 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 height of object ⇒ $ho$ height of image ⇒ $hi$ SOLUTION:- $ho=5cm\\f=-15cm\\u=-20cm\\\\using\:mirror \:formula\:we \\\:get\:position \:of\:image from\:mirror :-\\\\\frac{1}{f}= \frac{1}{v}+ \frac{1}{u}\\now \: put\:values\:in\:equation\\\frac{1}{-15}=\frac{1}{v}+\frac{1}{-20}\\\frac{1}{v}=\frac{-1}{15}+\frac{1}{20}\\\frac{1}{v}=\frac{-14+13}{60} \\\frac{1}{v}=\frac{-4+3}{60} \\\frac{1}{v}=\frac{-1}{60} \\v=-60cm\\\\put \:values\:in\:equation:-\\m=\frac{hi}{ho}=\frac{-v}{u} \\\frac{hi}{5}=\frac{-(-60)}{-50}\\hi=\frac{-605}{50}\\hi=-6cm$ 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 ( $hi$ ) ⇒ $-6cm$ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ 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.

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»

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

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.