Question

the focal length of concave lens is 50 cm if an object is kept at the distance of 75 cm from the lens then find the image distance and magnification produced by the lens​

Answer 1

Answer:

❣️ hello mate ❣️

Step-by-step explanation:

f=-50

u=-75cm

1/v-1/u=1/f

1/v =1/-50 +1/-75

it will be -5/150

=-1/30

image distance is 30cm in the same side

m= v/u

-30/-50

3/5times

hope you are satisfied

Answer 2

$\huge{\bf{\underline{\overline{\mathfrak{\red{Solution:-}\mid}}}}}$

We know that the focal length of a concave lens is always '-ve' and the image distance is also '-ve'

Here,

Focal length, f=-50cm

Image increases, u=-75cm

Now,

According to lens formula

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

$\implies\tt \frac{ 1}{ - 50} = \frac{ 1}{ v} - \frac{1}{-75}$

$\implies\tt \frac{1}{v} = \frac{1}{ - 50} + \frac{1}{( - 75)}$

$\implies\tt \frac{1}{v} = \frac{1}{ -50} - \frac{1}{75}$

$\implies\tt \frac{1}{v} = \frac{- 3 -2 \: \: }{ 150}$

$\implies\tt \frac{1}{v} = \frac{ -5 \: \: }{ 150}$

$\implies\tt v = - 30 \: cm$

So, the image distance is 30cm.

Again,

Magnification, m

$= \frac{ v }{u}$

$= \frac{ - ( 30)}{ - 75}$

$= \frac{ 2 }{5}$

$=.4$

Thus, the image is 0.4 times magnified.