Light reflection and refraction Worksheet-11

Light reflection and refraction Worksheet-11

1. Name the type of mirror used in the following situation also give reason:

Solar furnace.

2. One half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.

3. An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.

4. A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens ? Draw the ray diagram.

5. An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image ?

6. The magnification produced by a plane mirror is m = + 1. What does this mean ?

7. An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of image, its nature and size.

8. An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror, should a screen be placed, so that a sharp focused image can be obtained ? Find the size and nature of the image ?

9. Find the focal length of a lens of power –2.0 D. What type of lens is this ?

10. A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging ?

Answer:

1. For solar furnace, we use a concave mirror. Light from the sun, on reflection from the mirror, is concentrated at the focus of the mirror, producing heat.

2. Yes, it will produce a complete image of the object. To verify this  the image of a distance object like tree or pole on a screen, when lower half of the lens is covered with a black paper. The intensity or brightness of image will reduce.

3. Here, object size, h₁ = 5 cm

object distance, u = –25 cm

focal length of lens,  f = 10 cm

image distance, = ?

image size, h₂ = ?

As

∴ Putting the values

          16.67 cm.

As is positive, the image formed is real; on the right side of the lens.

As    

          ∴

Negative sign shows that the image is inverted.

4. Here, focal length of lens, f = –15 cm

image distance, v = –10 cm

object distance, u = ?

From        

∴

u = –30 cm. 

The ray diagram of image formation is shown in the figure.

5. Here, object distance, u = –10 cm focal length, f = 15 cm image distance, v = ?

As

∴

v = 6 cm

Here, + sign of v indicates that image is at the back of the mirror. It must be virtual, erect and smaller in size than the object.

6. As 

i.e., size of image is equal to size of the object. Further, + sign of m indicates that the image is erect and hence virtual.

7. Here, object size, h₁ = 5.0 cm

object distance, u = –20 cm

radius of curvature, R = 30 cm

image distance, v = ?

image size, h₂ = ?

As    

∴    

Positive sign of indicates that image is at the back of the mirror. It must be virtual and erect.

As    

∴    

          cm

This is the size of the erect image.

8. Here, object size, h₁ = 7.0 cm

object distance, u = –27 cm

focal length, f = - 18 cm

image distance, v = ?

image size, h₂ = ?

As    

∴    

v = –54 cm

∴ The screen should be held in front of the mirror at a distance of 54 cm from the mirror. The image obtained on the screen will be real.

As    

∴    

          h₂ = –14.0 cm

Negative sign of h₂ indicates that the image is inverted.

9. Here, focal length f = ?,  power P = –2.0 D

As     f =

∴ f = = –50 cm.

As power of lens is negative, the lens must be concave.

10. Given P = +1.5 D, f = ?

From f = , f = = 66.7 cm

The prescribed lens is converging or convex as its power is positive.