Question

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

Answer 1

Answer:

Given: convex lens

f=50cm

u=-75cm

v=?

m=?

By applying lens formula

1/f = 1/v-1/u

1/50=1/v - (-1/75)

1/50= 1/v +1/75

1/50-1/75=1/v

3-2/150=1/v

1/150 =1/v

150/1 =v

150cm=v

Magnification=v/u

m= 150/-75

m= -2

Explanation:

Answer 2

Answer:

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Explanation:

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

$⟹ \: \: \: \: \: \: \: \: \: \: \: \: \frac{1}{ƒ} = \frac{1}{v} - \frac{1}{4} \\ </u></strong></p><p></p><p><strong><u>[tex] ⟹ \: \: \: \: \: \: \: \: \: \: \: \: \frac{1}{ƒ} = \frac{1}{v} - \frac{1}{4}$

$⟹ \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{1}{ - 50} = \frac{1}{v} - \frac{1}{ - 75}$

$⟹ \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{1}{v} = \frac{ 1 }{ - 50} + \frac{1}{( - 75)}$

$⟹ \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{1}{v} = \frac{1}{ - 50} - \frac{1}{75}$

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

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

$⟹ \: \: \: \: \: \: \: \: \: \: \: \: \: v = - 30cm \:$

So, the image distance is 30cm.

Again,

Magnification, (m)

$\: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{v}{u}$

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

$\: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{2}{5}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: = 0.4$

Thus, the image is 0.4 times magnified