Question

An object of 8cm height is placed at a distance of 20cm from a convex lens of focal length 40cm. Find the position nature and size of the image

Answer 1

Answer:

$given$

OBJECT SIZE = 8CM

OBJECT DISTANCE = -20 CM (

$(object \: distance \: is \: always \: negative)$

FOCAL LENGTH = +40 CM

$(in \: convex \: lens \: focal \: length \: is \: positive)$

TO FIND :

POSITION, NATURE AND SIZE OF IMAGE. !!

FORMULA USED:

$\frac{1}{image \: distance} + \frac{1}{object \: distance} = \frac{1}{focal \: length}$

SO,

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

$\frac{1}{v} + \frac{1}{ - 20cm} = \frac{1}{40cm}$

$\frac{1}{v} = \frac{1}{40} - ( \frac{1}{ - 20} )$

$\frac{1}{v} = \frac{1}{40} + \frac{1}{20}$

$\frac{1}{v} = \frac{20 + 40}{40 \times 20}$

$\frac{1}{v} = \frac{60}{20 \times 40}$

$\frac{1}{v} = \frac{3}{40} cm$

$v = \frac{40}{3} cm$

IMAGE DISTANCE = 40/3 CM.

NOW,

SIZE OF THE IMAGE =

$\frac{ - image \: distance}{object \: distance} = \frac{height \: of \: image}{height \: of \: objevt}$

$\frac{ \frac{40}{3} }{ - 20} = \frac{h(i)}{8}$

$\frac{ - 2}{3} = \frac{h(i)}{8}$

$\frac{ - 16}{3} cm = height \: of \: image$

NATURE OF IMAGE :

1. MAGNIFIED

2. REAL AND INVERTED.

Plz Mark as brainliest....!!!!

Answer 2

QUESTION:

An object of 8cm height is placed at a distance of 20cm from a convex lens of focal length 40cm. Find the position nature and size of the image.

ANSWER:

SIGN COVENTION WHILE SOLVING OPTICS QUESTION :

SIGN COVENTION RELATED TO CONVEX LENS.

OBJECT DISTANCE

$(always \: negative)$

FOCAL LENGTH

$(always \: positive)$

Now come to main question :

GIVEN :

object size = 8 cm

object distance = -20 cm

Focal length = 40 cm

TO FIND :

Nature and size of the image?

__________________________________

By using mirror formula;

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

where;

$v = image \: distance \\ u = object \: distance \\ f = focal \: length$

so, by putting the given information in formula :

$\frac{1}{v} + \frac{1}{ - 20cm} = \frac{1}{40cm} \\ \frac{1}{v} - \frac{1}{20cm} = \frac{1}{40cm} \\ \frac{1}{v} = \frac{1}{40} + \frac{1}{20} \\ \frac{1}{v} = \frac{1 + 2}{40} cm \\ \frac{1}{v} = \frac{3}{40} cm \\ v = \frac{40}{3} cm$

(IMAGE DISTANCE = 40/3 cm )

Now, by using the magnification formula,

$magnification = \frac{ height \: of \: image}{height \: of \: image} = \frac{ - image \: distane}{object \: distance}$

$\frac{h(i)}{h(o)} = \frac{ \frac{ - 40}{3} }{ - 20} \\ \frac{h(i)}{8cm} = \frac{40}{3 \times 20} \\ \frac{h(i)}{8cm} = \frac{2}{3} \\ h(i) = \frac{2}{3} \times 8cm \\ h(i) = \frac{16}{3} cm$

(HEIGHT OF IMAGE = 16/3 CM )

NATURE OF IMAGE :

1. Magnified

$(as \: magnification \: is \: greater \: than \: 1)$

2. Real and inverted.

PLZ MARK AS BRAINLIEST.