a concave lens form and erect image of one third of size of the object

1.	a concave lens form and erect image of one third of size of the object which president at a distance of 30 cm in front of length find the position of the image and
Answer» ━━━━━━━━━━━━━━━━━━━━━━━━ ★correct Question★ a concave lens forms An erect IMAGE of 1/3 SIZE of the object which is placed at a DISTANCE 30 CM in front of a lens,find position of image, focal length of the lens ━━━━━━━━━━━━━━━━━━━━━━━━ Answer:- ★The focal length is 15 cm and the image is formed at 10 cm from the lens. Given:- $\bold{Distance\: of \:the\: object\: u \:= -30 \:cm}$ Magnification m = $\dfrac{h'}{h}= \dfrac{1}{3}$ We know that, The magnification is m = $\dfrac{h'}{h}=\dfrac{v}{u}$ $\dfrac{1}{3}=\dfrac{v}{-30}$ v = -10 cmv=−10cm The image is formed at 10 cm from the lens. Using lens's formula $\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$ $\dfrac{1}{f}=\dfrac{1}{-15}$ f = -15 cmf=−15cm Hence, The focal length is 15 cm and the image is formed at 10 cm from the lens. ━━━━━━━━━━━━━━━━━━━━━━━━

a concave lens form and erect image of one third of size of the object which president at a distance of 30 cm in front of length find the position of the image and

Answer»

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★correct Question★

a concave lens forms An erect IMAGE of 1/3 SIZE of the object which is placed at a DISTANCE 30 CM in front of a lens,find position of image, focal length of the lens

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Answer:-

★The focal length is 15 cm and the image is formed at 10 cm from the lens.

Given:-

$\bold{Distance\: of \:the\: object\: u \:= -30 \:cm}$

Magnification m = $\dfrac{h'}{h}= \dfrac{1}{3}$

We know that,

The magnification is

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

$\dfrac{1}{3}=\dfrac{v}{-30}$

v = -10 cmv=−10cm

The image is formed at 10 cm from the lens.

Using lens's formula

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

$\dfrac{1}{f}=\dfrac{1}{-15}$

f = -15 cmf=−15cm

Hence, The focal length is 15 cm and the image is formed at 10 cm from the lens.