Question

One half of a convex lens of focal length 10 cm is covered with a black paper. Can such a lens produce an image of a complete object placed at a distance of 30 cm from the lens ? Draw a ray diagram to justify your answer.
A 4 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 20 cm. The distance of the object from the lens is 15 cm. Find the nature, position and size of the image.

Answer 1

Answer:

Q1.

Yes but the intensity would be low. Refer to the attachment.

Q2.

Nature: Virtual and erect

Position: 60 cm in front of lens.

Size: 16 cm

Explanation:

Q2.

Given:

• Height of the object, ho = 4 cm
• Focal length, f = + 20 cm
• Object distance, u = - 15 cm

To find:

• Nature
• Position
• Size

Variables needed:

• Image distance, v = ?
• Magnification, m = ?
• Image height, hi = ?

Solution:

By using lens formula,

$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}\\\\\\\implies \dfrac{1}{20} = \dfrac{1}{v} - \dfrac{1}{-15}\\\\\\\implies \dfrac{1}{20} = \dfrac{1}{v} + \dfrac{1}{15}\\\\\\\implies \dfrac{1}{v} = \dfrac{1}{20} - \dfrac{1}{15} = \dfrac{15 - 20}{300} = \dfrac{-5}{300}\\\\\\\implies v = \dfrac{-300}{5} = -60\\\\\\\implies \boxed{\bold{v = - 60 \; \; cm}}$

Therefore, The image is formed at a distance of 60 cm in front of the lens.

Now,

$m = \dfrac{v}{u} = \dfrac{-60}{-15}\\\\\\\implies \boxed{\bold{m = 4}}$

Since m is +ve, The nature of the image is virtual and erect.

Also,

$m = \dfrac{hi}{ho}\\\\\\\implies 4 = \dfrac{hi}{4}\\\\\\\implies hi = 16\\\\\\\implies \boxed{\bold{hi = 16 \; \; cm}}$

Therefore, The height of image is 16 cm.