Question

calculate two possible distance of an object from a convex lens of focal length 20cm so as to obtain an image of double the size of an object​

Answer 1

Answer :

Given :

The focal length of the convex lens is 20 cm

To find :

the two possible distances of an object from the convex lens so as to obtain an image of double the size of an object​.

Solution :

The focal length of the convex lens, f = +20 cm

The size of the image is double the size of the object.

 hᵢ = 2hₒ

hᵢ/hₒ = 2

Magnification,

   $\underline{\boxed{\sf m=\dfrac{-v}{u}=\dfrac{h_i}{h_o}}}$

     

The two possible cases here are :

• formation of a real image
• formation of a virtual image

Real image :

The magnification is negative for real image produced by convex lens.

   m = -2

   -v/u = -2

     v = 2u

Using lens formula,

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

Substitute the values,

$\sf \dfrac{1}{20}=\dfrac{1}{2u}-\dfrac{1}{u} \\\\ \sf \dfrac{1}{20}=\dfrac{1-2}{2u} \\\\ \sf \dfrac{1}{20}=\dfrac{-1}{2u} \\\\ \sf 2u=-20 \\\\ \sf u=-20/2 \\\\ \sf u=-10 \ cm$

The object should be placed at 10 cm in front of the lens.

Virtual image :

The magnification is positive for virtual image produced by convex lens.

   m = +2

   -v/u = 2

     v = -2u

Using lens formula,

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

Substitute the values,

  $\sf \dfrac{1}{20}=\dfrac{1}{-2u}-\dfrac{1}{u} \\\\ \sf \dfrac{1}{20}=\dfrac{-1-2}{2u} \\\\ \sf \dfrac{1}{20}=\dfrac{-3}{2u} \\\\ \sf 2u=-60 \\\\ \sf u=-60/2 \\\\ \sf u=-30 \ cm$

The object should be placed at 30 cm in front of the lens.

Answer 2

Answer: The object should be placed at 10 cm in front of the lens for real image and object should be placed at 30 cm in front of the lens for virtual image.

Given: The focal length of the convex lens is 20 cm.

To find: The two possible distances of an object from the convex lens so as to obtain an image of double the size of an object​.

Step-by-step explanation:

The focal length of the convex lens is (f) = +20 cm

The size of the image is doubled the size of the object.

hᵢ = 2hₒ

hᵢ/hₒ = 2

Magnification,$m=\frac{-v}{u} =\frac{h_i}{h_0}$

There are only two possible cases-

1. Formation of a real image
2. Formation of a virtual image

Real image :

The magnification is negative for real image produced by convex lens.

m = -2

-v/u = -2

v = 2u

Using lens formula, $\frac{1}{f} = \frac{1}{v} -\frac{1}{u}$

Put the above values,

$\frac{1}{20} =\frac{1}{2u} -\frac{1}{u} \\\\\frac{1}{20} =\frac{1-2}{2u}\\\\\frac{1}{20} = \frac{-1}{2u} \\\\2u=-20\\\\u=\frac{-20}{2} \\\\u=-10 \ cm$

Hence, the object should be placed at 10 cm in front of the lens.

Virtual image :

The magnification is positive for virtual image produced by convex lens.

m = +2

-v/u = 2

v = -2u

Using lens formula, $\frac{1}{f} = \frac{1}{v} -\frac{1}{u}$

Substitute the values,

$\frac{1}{20} =\frac{1}{-2u} -\frac{1}{u} \\\\\frac{1}{20} =\frac{-1-2}{2u}\\\\\frac{1}{20} = \frac{-3}{2u} \\\\2u=-60\\\\u=\frac{-60}{2} \\\\u=-30 \ cm$

The object should be placed at 30 cm in front of the lens.

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