Question

A converging lens magnifies a real image to four times its original size. Given that the focal length of the converging lens is 20 cm. The object distance will be __________25cm
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Answer 1

Answer :-

→ Object Distance (u) = 25 cm

To Find :-

→ Object distance .

Explanation :-

According to the question ,

→ A converging ( convex) lens magnifies a real image to 4 times to its original size

We know that the magnification power

$\to \sf{m = \frac{v}{u} } \\ \\ \to \sf{ 4 = \frac{v}{u} } \\ \\ \to \boxed{\sf{v = 4u}}$

Now ,we have focal length of 20 cm , in convex lens "v" is taken positive .here u is negative so v = -4u hence

$\to \sf{ \frac{1}{f} = \frac{1}{v} + \frac{1}{ - u} } \\ \: \\ \to \sf{ \frac{1}{20} = \frac{1}{-4u} -\frac{1}{u} } \\ \: \\ \to \sf{ \frac{1}{20} =\frac{-(1+4)}{4u}} \\ \\ \to \sf{ \frac{1}{20} = \frac{ -5}{4u} } \\ \\ \to \sf{4u = -100} \\ \\ \to \sf{u = \frac{ -100}{4} } \\ \\ \to \boxed{\sf{ u = -25 \: cm \: }}$

object distance is negative hence ,our object is in left side .

→ Object Distance (u) = 25 cm

Answer 2

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QUESTION :

A converging lens magnifies a real image to four times its original size. Given that the focal length of the converging lens is 20 cm. The object distance will be Equal to 25cm

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SOLUTION :

From the question we can gather the following information :

F = 20 cm

R, THE RADIUS OF CURVATURE = 2 F = 40 cm .

The magnification of the image is 4 times.

=> v/u = 4

=> v = 4 u.

From the mirror Formulae,

1 / f = 1 / v - 1 / u

=> 1 / 20 = 1 / -4 u - 1 / u

=> 1 / 20 = -5 / 4u

=> u = - 25 cm.

So the object Distance becomes -25 cm...

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