A 5 cm tall object is placed perpendicular to the principle axis

Subject

Science

Class

CBSE Class 10

Pre Boards

Practice to excel and get familiar with the paper pattern and the type of questions. Check you answers with answer keys provided.

Sample Papers

Download the PDF Sample Papers Free for off line practice and view the Solutions online.
Advertisement

 Multiple Choice QuestionsShort Answer Type

21.

State one role of ciliary muscles in the human eye?

1414 Views

22. To construct ray diagram we use two light rays which are so chosen that it is easy to know their directions after reflection from the mirror. List these two rays and state the path of these rays after reflection. Use these rays to locate the image of an object placed between centre of curvature and focus of a concave mirror.
919 Views

23.

What is the colour of the sky during day time? Give reason for it?

504 Views

24.

Draw a labelled ray diagram to illustrate the dispersion of a narrow beam of white light when it passes through a glass prism.

1218 Views

Advertisement
Advertisement

25. A 5 cm tall object is placed perpendicular to the principle axis of a convex lens of focal length 12 cm. The distance of the object from the lens is 8 cm. Using the lens formula, find the position, size and nature of the image formed.


Given,

Object distance from the lens, u = - 8 cm

Focal length of the lens, f = 12 cm

Height of the object, h = 5 cm

Using the lens formula,

1 over straight v minus 1 over u equals 1 over f

Putting the values,

1 over straight v equals 1 over 12 plus fraction numerator negative 1 over denominator 8 end fraction
space space space space equals space 1 over 12 minus 1 over 8
space space space space equals space fraction numerator 2 minus 3 over denominator 24 end fraction
space space space space equals space fraction numerator negative 1 over denominator 24 end fraction

rightwards double arrow space v space equals space minus 24 space c m

Thus, image is formed 24 cm on the left of the convex lens.

That is, object is formed on the same side of the object.

Now,
Magnification, m = straight v over straight u equals fraction numerator h apostrophe over denominator h end fraction equals fraction numerator negative 24 over denominator negative 8 end fraction equals fraction numerator h apostrophe over denominator 5 end fraction
That is,

Height of the image, h' = 15 cm (positive)

Height of the image is 15 cm which is 3 times larger than the height of the object.

Therefore, the image formed is virtual, enlarged and erect.

795 Views

Advertisement
26.
State the types of mirrors used for: i) headlights and ii) rear view mirror, in cars and motorcycles. Give to justify your answer in each case.
1115 Views

27. An old man cannot see objects closer than 1 m from the eye clearly. Name the defect of vision he is suffering from. How can it be corrected? Draw ray diagram for the:

(i) defect of vision and also (ii) for its correction.
1697 Views

28.

Name two decomposers operating in our ecosystem.

686 Views

Advertisement
29.

Why is vegetative propagation practiced for growing some types of plant? List two plants which are grown by this method.

882 Views

30.

State the role of placenta in the development of embryo. 

488 Views

Advertisement