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Physics Part II

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Class 10 Class 12
Two thin lenses of power +4D and -2D are in contact. What is the focal length of the combination?

Power of first lens = +4 D
Power of second lens = -2 D
Power of the combination of lens,
P = P1+P2      = 4 - 2     = + 2D
Since focal length, f = 1P
Focal length of the lens,
 f = 12 = 0.5 m               = 50 cm
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Show that the least possible distance between an object and its real image in a convex lens is 4f where f is the focal length of the lens.


Suppose I is the real image of an object O. Let d be the distance between them. If the image distance is x, the object distance will be (d – x).
Thus,    u = – (d – x) and v = + x
Sustituting in the lens formula we have
                  1 over straight x minus fraction numerator 1 over denominator negative left parenthesis straight d minus straight x right parenthesis end fraction space equals 1 over straight f
or,                1 over straight x plus fraction numerator 1 over denominator left parenthesis straight d minus straight x right parenthesis end fraction equals space 1 over straight f
or,              straight x squared minus xd minus fd space equals 0

Suppose I is the real image of an object O. Let d be the distance bet
For a real image, the value of x must be real, i.e., the roots of the above equation must be real. This is possible if
d2 ≥ 4fd
or,    d ≥ 4f
Hence, 4f is the minimum distance between the object and its real image formed by a convex lens.

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An object 0.5 cm high is placed 30 cm from a convex mirror whose focal length is 20 cm. Find the position, size and nature of the image.

We have,
Distance of object from the mirror, u = –30 cm;
Distance of image, v = ?
Focal length, f = + 20 cm 

We know, using the mirror formula,
                     1v+1u =1f

                    v =ufu-f    =(-30) (+20)-30-20      = + 12 cm 

The image is virtual and erect as, it is formed at a distance of 12 cm behind the mirror. 

Now, size if the object, O = 0.5 cm 
sixe of the image = I
Magnification,  m = IO = -vu=-12-30
                             I = 25×O   = 25× 0.5  = +0.2 cm 

Hence, the height of the image = + 0.2 cm.
The positive sign indicates that the image is erect is smaller in size

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A convex lens of refractive index 1.5 has a focal length of 18 cm in air. Calculate the change in its focal length when it is immersed in water of refractive index 4/3. 

Given, a convex lens.
Refractive index of lens, μga  = 1.5 
Focal length of lens in air, fa = 18 
Refractive index of water, μwa = 43

For the lens in air,
               1fa = (μga-1) 1R1-1R2 

              118= (1.5 - 1) 1R1-1R2 

               1R1-1R2= 14 

When the lens is immersed in water
               1fw = μgaμwa-1 1R1-1R2 

                    = 1.54/3-1 × 14= 18× 14 = 132 

Thus,        fw = 32 cm.  

Hence, focal length changes from 18 to 32.

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An object 0.2 cm high is placed 15 cm from a concave mirror length 5 cm. Find the position and size of the image.

We have, 

object distance, u = -15 cmFocal length, f = -5 cmImage distance, v = ? 

Using formula, 

  1v+1u = 1f            

v = ufu-f        =(-15) (-5)-15+5        = -7.5 cm

The image is formed at a distance of 7.5 cm in front of mirror. 
Now,
Magnification, m = IO = -vu

where,
I is the image size and, 

O is the object size. 

             IO=-(-7.5)(-15)         I =-0.1 cm 

The negative sign indicates that the image is inverted and is diminished in size.

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A fish rising vertically to the surface of water in a lake uniformly at the rate of 3 m/s observes a king fisher bird diving vertically towards water at the rate 9 m/s vertically above it. If the refractive index of water is 4/3, find the actual velocity of the dive of the bird.

If at any instant, the fish is at a depth ‘x’ below water surface while the bird at a height y above the surface, then the apparent height of the bird from the surface as seen by the fish will be given by
straight mu space equals space fraction numerator Apparent space height over denominator Real space height end fraction

If at any instant, the fish is at a depth ‘x’ below water surface
or Apparent height = μy
So, the total apparent distance of the bird as seen by the fish in water will be h = x + μy
or,                      dh over dt equals dx over dt plus straight mu dy over dt
or,                        9 space equals space 3 plus straight mu open parentheses dy over dt close parentheses
or,                        dy over dt space equals space fraction numerator 6 over denominator left parenthesis 4 divided by 3 right parenthesis end fraction space equals space 4.5 space straight m divided by straight s.
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