Question

➡️There are two lens of focal lengths 50cm and -25cm. An object is kept at a distance of 100cm from each of the lenses. Calculate: (1) Image distanice
(2) magnification in each of the two cases.

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Answer 1

Explanation:

(a) PA = + 2D

fA = 1/PA = 1/2 = + 0.5 m = +50 cm

Lens A is a convex lens.

PB = -4D

fB = 1/PB = 1/-4 = -0.25 m = -25 cm

Lens B is a concave lens.

(b) case 1 : for lens A

fA = +50 cm

uA = -100 cm

1/fA = 1/VA – 1/UA

1/VA = 1/f A + 1/uA

= 1/50 + 1/-100

= 1/100

Image distance, VA = 100 cm

Magnification, m A = vA/uA = 100/-100 = -1

Case 1: For lens B

fB = - 25 cm

uB = -100 cm

1/fB = 1/vB + 1/uB

1/vB = 1/fB + 1/uB

= 1/-25 + 1/-100

= -5/100

Image distance, VB = -20 cm

Magnification, mB = vB/uB = -20/-100 = +0.2

Answer 2

Given:

There are two lens of focal lengths 50cm and -25cm. An object is kept at a distance of 100cm from each of the lens.

To find:

(1) Image distance

(2) magnification in each of the two cases.

Calculation:

1st case:

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

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

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

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

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

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

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

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

$\sf \implies \: magnification= \dfrac{100}{( - 100)}$

$\sf \implies \: magnification= - 1$

So, image distance is 100 cm and magnification is -1.

2nd case:

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

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

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

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

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

$\sf \implies \: \dfrac{1}{v} = \dfrac{ - 3}{100}$

$\sf \implies \: v = - 33.3 \: cm$

$\sf \implies \: magnification = \dfrac{ - 33.3}{ - 100}$

$\sf \implies \: magnification = 0.33$

So, image distance is 33.3 cm and magnification is +0.33.