Question

11. The image obtained with a converging lens is upright and three times the length of the object,
if the focal length of the Irns is 20cm.Calculate the object distance
(A) 13.3cm (B) 40cm (C) 33.1cm (D) 40.3cm (E) 34.9cm

Answer 1

Given :

The image obtained by convex lens is three times the length of the object.

Focal length = 20cm

To Find :

Distance of object from the lens.

Solution :

■ First of all we need to derive relation between distance of object and distance of image.

Let's apply formula of magnification$.$

➠ m = v/u

➠ 3 = v/u

➠ v = 3u

✵ Focal length of convex lens is taken positive and that of concave lens is taken negative.

Distance of object can be calculated by using lens formula which us given by

⭆ 1/v - 1/u = 1/f

⭆ 1/(3u) - 1/u = 1/20

⭆ (1 - 3)/3u = 1/20

⭆ -3u/2 = 20

⭆ -3u = 40

⭆ u = -40/3

⭆ u = -13.3 cm

∴ (A) is the correct answer!

Answer 2

Given :-

• hi =+ 3ho( upright , virtual )
• focal length = +20 cm (converging lens - convex lens

Solution :-

Magnification of a lens is given by ,

$\longrightarrow\boxed{\mathtt{m = \dfrac{v}{u} = \dfrac{h_i}{h_o}}}$

here ,

• m = magnification
• v = image distance
• u = object distance
• hi = height of image
• ho = height of object

hence ,

$\longrightarrow\mathtt{ v = \dfrac{h_i \times u}{h_o}}$

$\longrightarrow\mathtt{ v = \dfrac{3\not{h_o}\times u}{\not{h_o}}}$

$\longrightarrow\mathtt{ v = 3u}$

As per the lens formula ,

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

here ,

• f = focal length
• v = image distance
• u = object distance

$\longrightarrow\mathtt{ \dfrac{1}{f} = \dfrac{1 }{3u}- \dfrac{1}{u}}$

$\longrightarrow\mathtt{ \dfrac{1}{f} = \dfrac{1 - 3}{3u}}$

$\longrightarrow\mathtt{ \dfrac{1}{f} = \dfrac{-2}{3u}}$

$\longrightarrow\mathtt{ u = \dfrac{-2f}{3}}$

Substituting the value of f in the above equation we get ,

$\longrightarrow\mathtt{ u = \dfrac{-2 x 20}{3}}$

$\longrightarrow\mathtt{ u = \dfrac{-40}{3}}$

$\longrightarrow\boxed{\mathtt{ u = - 13.33 cm }}$

The value of image distance is negative as distances are always measured from the optical center of the lens

The object is placed at a distance of 13.33 cm in front of the lens