Light reflection and refraction Worksheet9

An object is placed at a distance or 50 cm from a concave lens of focal length 30 cm. Find the nature and position of the image.

An object of height 2 cm is placed at a distance of 15 cm in front of a concave lens of power –10 dioptre. Find the size of the image.

A thin lens has a focal length of –25 cm. What is the power of the lens ? Is it convex or concave ?

The power of lens is 25 D. What is its focal length?

A convergent lens of power 8 D is combined with a divergent lens of power –10 D. Calculate focal length of the combination.

A concave lens is kept in contact with a convex lens of focal length 20 cm. The combination works as a converging lens of focal length 100 cm. Calculate power of concave lens.

Find the focal length and nature of lens which should be placed contact with a lens of focal length 10 cm so that the power of the combination becomes 5 dioptre.

The radius of curvature of a spherical mirror is 20 cm. What is its focal length ?

Name a mirror that can give an erect and enlarged image of an object.

Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Answer:

Here, = –50 cm, f = –30 cm, v = ?
As
–18.7 cm
As v is negative, the image must be virtual and erect.

Here, h_{1} = 2cm, u = –15 cm
P = –10D, h_{2} = ?
Now,
As
∴
As v is negative, image is virtual
As
∴
h_{2} = 0.8 cm. As h_{2} is positive, image is erect.

Here,
From
Negative sign shows that lens is concave.

Here,
∴
= 40 cm.

Here, P_{1} = 8 D, P_{2} = –10 D, f = ?
As P = P_{1} + P_{2} = 8 – 10 = –2D
∴ –0.5 m

Here, f_{1} = 20 cm,
F = 100 cm,
As P_{1} + P_{2} = P, ∴ P_{2} = P – P_{1} = 1 – 5 = –4 D

Here, f_{1} = ?, f_{2} = 10 cm,
P = 5D.
As P_{1} + P_{2} = P
∴ P_{1} + 10 = 5 or P_{1} = 5 – 10 = – 5D.
∴ –20 cm.
Lens must be concave.

Here, R = 20 cm, f = ?
As ∴

A concave mirror gives an erect and enlarged image of an object held between pole and principal focus of the mirror.

Here, focal length, f = ?
radius of curvature,
As