Question

There are two lens of focal lengths 50cm and-25cm
. An object is kept at a distance of 100 cm from each of the lenses.calculate
image distance and
magnification in each of the two cases
​

Answer 1

Answer:

Explanation:

Part - 1

Given,

Focal length, f = 50 cm

Object distance, u = - 100 cm

To Find,

Image distance, v ,and magnification, m

Formula to be used,

Lens formula, 1/f = 1/v - 1/u

Magnification formula, m = v/u = h'/h

Solution,

Putting all the values, we get

1/f = 1/v - 1/u

⇒ 1/50 = 1/v - 1/- 100

⇒ 1/50 = 1/v + 1/100

⇒ 1/v = 1/50 - 1/100

⇒ 1/v = 2 - 1/100

⇒ 1/v = 1/100

⇒ v = 100 cm

Hence, the image distance is 100 cm.

Now, the magnification,

Magnification formula, m = v/u

⇒ m = v/u

⇒ m = 100/- 100

⇒ m = - 1

Hence, the magnification is - 1.

Part - 2

Given,

Focal length, f = - 25 cm

Object distance, u = - 100 cm

To Find,

Image distance, v ,and magnification, m

Formula to be used,

Lens formula, 1/f = 1/v - 1/u

Magnification formula, m = v/u = h'/h

Solution,

Putting all the values, we get

1/f = 1/v - 1/u

⇒ 1/- 25 = 1/v - 1/- 100

⇒ 1/- 25 = 1/v + 1/100

⇒ 1/v = 1/- 25 - 1/100

⇒ 1/v = - 4 - 1/100

⇒ 1/v =  - 5/100

⇒ v = - 20 cm

Hence, the image distance is - 20 cm.

Now, the magnification,

Magnification formula, m = v/u

⇒ m = v/u

⇒ m = - 20/- 100

⇒ m = 0.2

Hence, the magnification is 0.2.

Answer 2

Given:-

• There are two lens of focal lengths 50cm and -25cm.

• An object is kept at a distance of 100 cm from each of the lenses.

To Find:-

• Image distance and magnification in each of the two cases.

Formula Used:-

• ${\boxed{\bf{Lens\: Formula: \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}}}}$

• ${\boxed{\bf{Magnification\: Formula: m=\dfrac{v}{u}=\dfrac{h_{(i)}}{h_{(o)}}}}}$

Here,

• v = Distance of image

• u = Distance of object

• f = Focal length

• m = Magnification

• $\bf h_{(i)}$ = Height of Image

• $\bf h_{(o)}$ = Height of Object

Solution:-

For 1st Lens(50cm):-

Using Lens Formula,

$\sf :\implies\: \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Here,

• f = 50cm

• u = -100

Putting values,

$\sf :\implies\: \dfrac{1}{50}=\dfrac{1}{v}-\dfrac{1}{-100}$

$\sf :\implies\: \dfrac{1}{50}=\dfrac{1}{v}+\dfrac{1}{100}$

$\sf :\implies\: \dfrac{1}{50}-\dfrac{1}{100}=\dfrac{1}{v}$

$\sf :\implies\: \dfrac{1}{v}=\dfrac{2-1}{100}$

$\sf :\implies\: \dfrac{1}{v}=\dfrac{1}{100}$

$\bf :\implies\: v=100\:cm$

Hence, The distance of image from the lens of focal length 50cm is 100cm.

Now, Using Magnification Formula,

$\sf :\implies\: m=\dfrac{v}{u}$

$\sf :\implies\: m=\dfrac{100}{-100}$

$\bf :\implies\: m=-1$

Hence, The Magnification is -1 which means the object is diminished.

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For 2nd Lens(-25cm):-

Using Lens Formula,

$\sf :\implies\: \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Here,

• f = -25cm

• u = -100

Putting values,

$\sf :\implies\: \dfrac{1}{-25}=\dfrac{1}{v}-\dfrac{1}{-100}$

$\sf :\implies\: \dfrac{1}{-25}=\dfrac{1}{v}+\dfrac{1}{100}$

$\sf :\implies\: \dfrac{1}{-25}-\dfrac{1}{100}=\dfrac{1}{v}$

$\sf :\implies\: \dfrac{1}{v}=\dfrac{-4-1}{100}$

$\sf :\implies\: \dfrac{1}{v}=\dfrac{-5}{100}$

$\bf :\implies\: v=-20\:cm$

Hence, The distance of image from the lens of focal length 25cm is 20cm.

Now, Using Magnification Formula,

$\sf :\implies\: m=\dfrac{-20}{-100}$

$\sf :\implies\: m=\dfrac{2}{10}$

$\bf :\implies\: m=0.2$

Hence, The Magnification is 0.2 which means the object is magnified.

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