Question

An object is placed at a distance of 15cm from a concave
lens of focal length 6cm. Find the location, nature and
size of the image.​

Answer 1

Answer:

Hope it helps you friend.

Answer 2

v=-4.3 cm

Nature:Virtual and erect

Size: Diminished

Step-by-step explanation:

Distance of object from optical center=u=-15 cm

Focal length of concave lens=f=-6 cm

Focal length of concave lens is always negaitive.

Lens formula

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

Where f=Focal length of

v=Image distance

u=Object distance

Substitute the values

$\frac{1}{-6}=\frac{1}{v}-\frac{1}{-15}=\frac{1}{v}+\frac{1}{15}$

$\frac{1}{v}=-\frac{1}{6}-\frac{1}{15}$

$\frac{1}{v}=\frac{-5-2}{30}$

$\frac{1}{v}=\frac{-7}{30}=-\frac{7}{30}$

$v=-\frac{30}{7}=-4.3cm$

Magnification=$\mid \frac{v}{u}\mid =\mid{-\frac{-4.3}{-15}}\mid =0.29<1$

The magnification is less than 1 .Therefore, the image is diminished.

Virtual and erect image is formed in concave lens.

It is formed on the side of object .

#Learns more:

https://brainly.in/question/2586152:Answered by Aviralmehrotrap