Question

when an object of height 6 cm is placed at a distance of 12 cm from the focal centre of the lens the objects 12 cm height reflection is form so,

1. indicate the type of lens?
2. find the magnification of reflection?
3. What is power of lens?​

Answer 1

Answer:(1)Convex

(2)2

(3)0.04167

Explanation:m=h'/h=v/u

12/6=v/-12

v=-12×12/6=-24cm

(1)it is a convex lens because it forms a bigger image of the object.

(2)m=h'/h=12/6=2

(3)1/v-1/u=1/f

1/f=(1/-24)-(1/-12)=1/24

Hence, power =1/f=1/24=0.04167(approx)

Answer 2

(a). The lens is converging.

(b). The magnification is 2.

(c). The power is 4.16 D.

Explanation:

Given that,

Height of the object = 6 cm

Height of the image = 12 cm

Object distance = 12 cm

We need to calculate the image distance

Using formula of magnification

$m=\dfrac{v}{u}=\dfrac{h'}{h}$

Put the value into the formula

$\dfrac{v}{-12}=\dfrac{12}{6}$

$v=-24\ cm$

We need to calculate the magnification

Using formula of magnification

$m=\dfrac{v}{u}$

Put the value into the formula

$m=\dfrac{-24}{-12}$

$m=2$

We need to calculate the focal length

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{f}=\dfrac{1}{-24}+\dfrac{1}{12}$

$f=24\ cm$

The focal length is positive so the lens will be converging.

We need to calculate the power of lens

Using formula of power

$P=\dfrac{100}{f}$

Put the value into the formula

$P=\dfrac{100}{24}$

$P=4.16\ D$

Hence, (a). The lens is converging.

(b). The magnification is 2.

(c). The power is 4.16 D.

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Topic : optics

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