Monday, October 28, 2013

CBSE I NCERT 10th Numerical Problem solved For _ Reflection and reflection of light


Q. 1. A concave mirror of focal length 20cm is placed 50 cm from a wall. How far from the wall an object be placed to form its real image on the wall?

Soluion : V=-50 cm    F=-20cm

From mirror formula 1/u = 1/f – 1/v

= -1/20+ 1/50 =-3/100      U = -33.3 cm

Therefore the distance of the object from the wall x = 50 – u 

X = 50 – 33.3 = 16.7 cm.

Q.2. An object is placed at a distance of 40cm from a concave mirror of focal length 15cm. If the object is displaced through a distance of 20 cm towards the mirror, By how much distance is the image displaced?


Answer:  Here f = -15 cm, u = -40 cm

Now 1/f = 1/u + 1/v

Then 1/v = 1/f – 1/u

Or   V= uf/u-f = -40 X -15/25 = -24 cm

Then object is displaced towards the mirror let u1 be the distance object from the

Mirror in its new position.

Then u1 = -(40-20) = -20cm

If the image is formed at a distance u1 from the mirror then

v= u1f/u1-f = -20X-15/-20+15 = -60 cm.  =  - 20 x-15/-20+15 = -60 cm. 

Therefore the image will move away from the concave mirror through a distance equal to 60 – 24 = 36 cm.

Q.3. An object is placed at distance of 25 cm from a spherical mirror and its image is formed behind the mirror at distance of 5 cm. Find focal length? Is it concave or convex mirror? Answer:
Here u = -25 cm , v = 5 cm from the mirror formula 1/f =1/u + 1/v

Then 1/f = -1/25 + 1/5 = 4/ 25

F = 6.25 cm

As the focal length is positive the mirror is convex in nature. 


Q. An object is placed in front of a convex mirror of radius of curvature 40 cm at a distance of 10 cm. Find the position, nature and magnification of mirror.
Answer:  
Here u = -10 cm, R = -40 cm

Then f = R/2 = - 20 cm

From the mirror formula 1/v = 1/f – 1/ u = -1/20 + 1/10 = 1/20.

V= 20 cm so v is positive , a virtual and erect image will be formed on the other side of the object, i.e; behind the mirror.  M=-v/u = -20/-10 = 2 

Q. An object is kept in front of a concave mirror of focal length of 15 cm. the image formed is 3 times the size of the object. Calculate the two possible distances of the object from the mirror. 

Answer: 
Case:1. Image is real. M = -3

Here f= -15 cm

Now m=-v/u = -3

Or , V = 3u

From the mirror formula

1/f = 1/u+1/v

-1/15 = 1/u + 1/3u

U= - 20 cm.

Case:2. When the image is virtual m = 3

Now m = -v/u = 3

Or,    V=-3
From the mirror formula

1/f=1/u+1/v

Then -1/15 = 1/u-1/3u

2/3u = -1/15

U=-10cm.

Q. Refractive index of glass is 1.5 and that of water is 1.3. if the speed of light in water is 2.25X108m/s. What is the speed of light in glass?
Answer: 
 Here ng=1.5 and nw=1.3 let v1 and v2 be the speeds of light in glass and water respectively. If c is the speed of light in air then

c/v1=1.5 and c/v2=1.3

then v1/v2= 1.3/1.5

v1=1.3/1.5X2.25X108

v1=1.95X108 m/s

Q. A light of wave length 6000A0 in air enters a medium with refractive index 1.5. what will be the frequency and wave length of light in medium?

Answer: 
Here wave length of light in air l=6000A0 = 6X10-7m

Refractive index of medium = 1.5 = n

The frequency of light does not change, when light travels from air to a refracting medium.

n=c/l = 3X108/6X10-7 = 5X1014Hz

The wave length of light in the medium l1=l/n = 6000/1.5 = 4000A0

Q. Convex lens made up of glass of refractive index 1.5 is dipped in turn in 
(i) Medium A of n=1.65
 ii) Medium B of n = 1.33. 
Explain giving reasons Whether it will behave as a converging lens or diverging lens in each of the

Answer: 

Two cases.
Here ng
Let fair be the focal length of lens in air then
   1/fair = (ng-1)(1/R1-1/R2)
 (1/R1-1/R2) = 1/fair(ng-1) = 2/fair……………………………..(1)

(i)  When lens is dipped in a medium A here nA = 1.65

Focal length be fA when dipped in a medium A then 1/fA=(nA-1)(1/R1-1/R2)

Using equation (1) we have

1/fA = (1.5/1.6 5-1) X 2/fair = -1/5.5 fair

FA = -5.5 fair

As sign of fA is opposite to that of fair, the lens will behave as diverging lense.

(ii)   When lens is dipped in a Medium B nB = 1.33

Let fB be the focal length of lens when dipped in medium B

Then 1/fB = (ng-1)(1/R1-1/R2) =(ng/nB-1) )(1/R1-1/R2)

1/fB =(1.5/1.33-1)X2/fair = 0.34/1.33 fair = F= 3.91

fair as the Sign of fB is same as that of fair the lens will behave as a converging lens

Q. A convergent beam of light passes through a diverging lens of focal length 0.2 meters comes to focus at a distance 0.3 meters behind the lens find the position of the point at which the beam would converge in the absence of lens?

Answer: F=-0.2 m, v=0.3 m

From the lens equation 1/f = 1/v-1/u

1/u = 1/v-1/f = 1/0.3-1/-0.2 = 50/6

U=6/50 = 0.12 m

In the absence of lens the beam would converge at a distance 0.12 m from the present position of the lens.

Q. A beam of light converges to a point P. A lens is placed in the path of convergent beam 12 cm from the point P. At what point the beam converges if the lens is   (a) a concave lens of focal length 16 cm  (b) a convex length of focal length 20 cm.

Answer: (a) here u =12cm f = -16 cm the lens equation we have

V=uf/u+f=12X-16/12-16 = 48 cm 

as v is positive the beam converges on the same side that of point P

b) Here u = 12 cm , f =20 cm from the lens equation we have

V=uf/u+f = 12X20/12+20 = 240/32 = 7.5 cm

As  v is positive the beam converges on the same side as that of point of P

Q.A converging and a diverging of equal focal lengths are placed co-axially in  Contact. Find the focal length and power of the combination.

Answer: Let f and –f are be the focal length of the converging and diverging lens respectively then focal length of the combination

1/F = 1/f – 1/f = 0

Power of the combination p = 1/F = 0.

Practice Questions

1. An object is placed in front of a concave mirror of radius of curvature 15cm at a distance of (a) 10cm. and (b) 5cm. Find the position, nature and magnification of the image in each case.

2. An object is placed 15cm from a concave mirror of radius of curvature 60 cm. Find the position of image and its magnification?

3. An object is kept at a distance of 5cm in front of a convex mirror of focal length 10cm. Give the position, magnification and the nature of the image formed.

4. An object is placed at a distance of 50cmfrom a concave lens of focal length 20cm. Find the nature and position of the image.

5. The power of a lens is 2.5 dioptre. What is the focal length and the type of lens?

6. What is the power of a concave lens of focal length 50cm?

7. Find the power of a concave lens of focal length 2m.

8. Two lens of power +3.5D and -2.5D are placed in contact. find the power and focal length of the lens combination.

9. A convex lens has a focal length of 20 cm. Calculate at what distance from the lens should an object be placed so that it forms an image at a distance of 40cm on the other side of the lens. State the nature of the image formed?

10. A 10cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 30cm. The distance of the object from the line is 20cm.find the i)position ii)nature and iii) size of the image formed.

11. Find the focal length of a line power is given as +2.0D.

12. With respect to air the refractive index of ice and rock salt benzene are 1.31 and 1.54 respectively. Calculate the refractive index of rock salt with respect to ice.

13. An object 5cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30cm. Find the position of the image, its nature and size.

14. The far point of a myopic person is 150cm in front of the eye. Calculate the focal length and the power of the lens required to enable him to see distant objects clearly.
Links For   CBSE Class X Science Term-II

2 comments:

J Sunil said...

The far point of a myopic person is 150 cm in front of the eye. Calculate the focal length and the power of a lens required to enable him to see distant objects clearly.

Solution:
u = infinity
v = -150cm
1/f = 1/v - 1/u
f = -150cm = -1.5m
P = 1/-1.5 = -10/15 = -2/3 = -0.66 D

J Sunil said...

An object 5 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 the image, its nature and size.

Soln.: Given h = 5 cm, u =-20 cm, R = +30 cm or f = R/2 = 30 /2 = +15 cm, v = ? h = ?

Using 1/f = 1/v - 1/u
we get v = 8.6 cm
The image is formed 8.6 cm behind the mirror. Thus the image is virtual and erect.

Now m = -v/u =hi/ho
hi = -v/u x ho = -8.6/-20 x 5 = 2.15 cm

Thus the size of the image is 2.15 cm or 2.2. cm. The image is reduced.

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