Question

51. A converging lens forms a five fold magnified image
of an object. The screen is moved towards the object
by a distance 0.5m, and the lens is shifted so that the
image has the same size as the object. Find the lens
power and the initial distance between the object
and the screen.
Ans: 6.4D. 1.125m]​

Answer 1

Answer:

6.4D. 1.125m

Explanation:

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Answer 2

The power of the lens is 6.4 D.

The initial distance between the object and the screen is 1.125 m

Explanation:

Given that,

Magnification = 5

Distance = 0.5 m

According to question,

The equation is formed,

$6x=2y+d$

Put the value in the equation

$6x-2y=0.5$...(I)

Let the object distance is x m and image distance is 5x m.

In first case,

We need to calculate the image distance

Using formula of magnification

$m=\dfrac{v}{u}$

Put the value into the formula

$5=\dfrac{v}{u}$

$v=5u$

We need to calculate the focal length

Using formula of lens

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

Put the value into the formula

$\dfrac{1}{f}=\dfrac{1}{5x}+\dfrac{1}{x}$

$\dfrac{1}{f}=\dfrac{6}{5x}$

$f=\dfrac{6f}{5}$....(II)

In second case,

$v=u$

Let the object distance is y m and image distance is y m.

We need to calculat the focal length

Using formula of lens again

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

Put the value into the formula

$\dfrac{1}{f}=\dfrac{1}{y}+\dfrac{1}{y}$

$\dfrac{1}{f}=\dfrac{2}{y}$

$y=2f$.....(III)

Put the value of x and y in eqution (I)

$6\times\dfrac{6f}{5}-2\times2f=0.5$

$f=\dfrac{2.5}{16}$

$f=0.156\ m$

We need to calculate the power of the lens

Using formula of power

$P=\dfrac{1}{f}$

Put the value of f

$P=\dfrac{1}{0.156}$

$P=6.4\ D$

We need to calculate the value of x

Using equation (I)

$x=\dfrac{6f}{5}$

Put the value of f

$x=\dfrac{6\times0.156}{5}$

$x=0.1872\ m$

The initial distance between the object  and the screen is

$d' =6x$

Put the value into the formula

$d'=6\times0.1872$

$d'=1.125\ m$

Hence, The power of the lens is 6.4 D.

The initial distance between the object and the screen is 1.125 m

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Topic : optics

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