Question

A convex lens forms an image 16.0 cm long of an object 4.0 cm long kept at a distance of 6 cm from the lens. The object and the image are on the same side of the lens.
(a) What is nature of the image?
(b) Find (i) the position of the image, and (ii) the focal length of the lens.​

Answer 1

Answer:

Given Information,

Height of the image = 16 cm

Height of the object = 4 cm

Object distance (u) = - 6 cm (Sign Convention)

To Find,

Nature of Image

Image Distance (v)

Focal Length of Lens

Solution,

Magnification of an object placed in front of a lens is given as:

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

Substituting the given information in the above formula we get:

$\implies \dfrac{h_i}{h_o} = \dfrac{v}{u}\\\\\\\implies \dfrac{16}{4} = \dfrac{v}{-6}\\\\\\\implies 4 = \dfrac{v}{-6}\\\\\\\implies v = -6 \times 4\\\\\implies \boxed{ \bf{ v = -24\:\:cm}}$

Hence the image distance is 24 cm towards the left side of the optic center. (Negative sign is due to the sign convention).

Since Magnification is positive (+4), the image is virtual. Also since the magnitude (4) is greater than 1, the image is magnified. Hence the image is virtual, erect and magnified.

Focal Length of the Lens can be calculated by the formula:

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

Substituting the values along with the sign convention, we get:

$\implies \dfrac{1}{f} = \dfrac{1}{-24} - \dfrac{1}{(-6)}\\\\\\\implies \dfrac{1}{f} = \dfrac{1}{-24} + \dfrac{1}{(6)}\\\\\\\text{Taking LCM we get:}\\\\\implies \dfrac{1}{f} = \dfrac{-1+4}{24} = \dfrac{3}{24}\\\\\\\implies \dfrac{1}{f} = \dfrac{1}{8}\\\\\\\implies \boxed{ \bf{ f = 8\:\:cm}}$

Hence the focal length of the lens is 8 cm.

Answer 2

Answer:

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