20 October 2012

10th Human Eyes ,Reflection and Refraction Numerical CBSE Test Series

10th Human Eyes ,Reflection and refraction Numerical CBSE Test Series

1. An object 5.0 cm in length is placed at a distance of 10cm from a convex mirror of radius of curvature 150cm. Find the position, nature and size of the image.

2.Find the position, nature and size of image of an object 4cm high placed at a distance of 10 cm from a concave mirror of focal length 20 cm.

3.An object is placed at a distance of 25 cm from the pole of a spherical mirror which forms a real, inverted image on the same side of object at 37.5 cm from the pole. Calculate the focal length of mirror and find nature of the mirror.

4.An object 2.0 cm in size is placed 20.0 cm in front of a concave mirror of focal length 10.0 cm. Find the distance from the mirror at which a screen should be placed in order to obtain a sharp image. What will be the size and nature of the image formed?

5.Find the position of an object which when placed in front of a concave mirror of focal length 20 cm produces a virtual image, which is twice the size of the object.

6.A concave lens has a focal length of 15cm. At what distance should an object 10 cm long be placed so that it forms an image of 10 cm from the lens? Find the nature and size of the image formed.

7.A convex lens has a focal length of 30 cm. Calculate at what distance should the object be placed from the lens so that it forms an image at 60 cm on the other side of the lens. Find the magnification produced by the lens in this case.
8.A concave lens has focal length 20cm. At what distance from the lens a 5 cm tall object be placed so that it forms an image at 15 cm from the lens? Also calculate the size of the image formed.
9.A Calculate the focal length of convex lens which produces a virtual image at a distance of 25 cm of an object placed 10cm in front of it.

10.A concave lens of focal length 15 cm forms an image of 10 cm from the lens. How far is object from the lens? What are its characteristics?

11.Light enters from air into glass plate which has a refractive index of 1.5. Calculate the speed of light in glass. (Given, speed of light in vacuum is 3 x 108ms-1
12.A person cannot see distinctly any object placed beyond 40 cm from his eye. Calculate the power of the lens which will help him to see distant object clearly.

13. The near point of a hypermetropic person is 75 cm from the eye. What is the power of lens required to enable him to read clearly a book held at 25 cm from the eye?

14.A person with a myopic eye is not able to see beyond 3 m distinctly. Determine the nature, focal length and power of the lens required.

15.The near point of a hypermetropic person is 1m. What is the power of lens required to correct this defect? Assume that near point of the normal eye is 25 cm.
10th Light Reflection and refractions : Related Posts Links

4 comments:

J Sunil said...

Q. If an object of height 4 cm is placed at distance of 12 cm from a concave mirror having focal length 24 cm, find the position, nature and the height of the image.

SOLUTION

Here,

object height h = 4 cm

object dis tance u = -12 cm

focal length f = -24 cm

Putting these values in mirror formula,

(1/u) + (1/v) = (1/f)

∴ (-1/12) + (1/v) = (-1/24)

∴ (1/v) = (1/12) - (1/24)

∴ (1/v) = (1/24)

∴ v = 24

Now, magnification m = (-u/v) = (-24)/(-12) = 2

∴ (h'/h) = 2

∴ h' = 2h = 2 x 4 = 8 cm

Since v is positive, the image is formed behind the mirror at the distance of 24 cm from the mirror.It is virtual and has height of 8 cm.

J Sunil said...

Q. An object of height 6 cm is placed at a distance of 10 cm from a convex mirror with radius of curvature 30 cm.Find the position, nature and the height of its image.

SOLUTION

Here,
Object distance u = -10 cm
Radius of curvature R = 30 cm
Object height h = 6 cm



Now,

focal length f = (R/2) = 30/2 = +15

Using mirror formula,

(1/u) + (1/v) = (1/f)

∴ (1/v) = (1/f) - (1/u)

∴ (1/v) = (1/15) - (1/-10)

∴ (1/v) = (1/15) + (1/10) =(1/6)

∴ v = 6 cm

Also, magnification m = -(u/v) = -(6/-10) = 0.6

∴ (h'/h) = 0.6

∴ h' = mh =0.6 x 6 = 3.6 cm

Thus, the image is formed behind the mirror(as v is positive), it is virtual and its height is 3.6 cm.

J Sunil said...

Q. At what distance the object should be placed so that the image will be formed at a distance 10 cm from a concave lens ? Focal length of the lens is 20 cm.

SOLUTION

Since the lens is concave, the image will be formed on the same side as that of the object.

Here,

Image distance v = -10 cm

Focal length f = -20 cm

Object distance u = ?

Using lens formula,

(1/f) = (1/v) - (1/u)

∴ (1/u) = (1/v) - (1/f)

∴ (1/u) = (1/-10) - (1/-20)

∴ (1/u) = (-2 + 1)/20

∴ (1/u) = -(1/20)

∴ u = -20 cm

Thus the object should be placed 20 cm from the concave lens on the left side of the lens.

J Sunil said...

Q.A ray of light enters from water to glass.Refractive index of glass with respect to water is 1.12. Find absolute refractive index of water if absolute refractive index of glass is 1.5.

SOLUTION

Taking water as "medium1" and glass as "medium2",

Absolute refractive index of water, η1 = ?

Absolute refractive index of glass, η2 = 1.5

η21 = 1.12_________(given)

∴ η2/η1 = 1.12

∴ η1 =η2/1.12

∴ η1 = (1.5)/(1.12) = 1.34

Hence,absolute refractive index of water = 1.34

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