Physics, asked by Anonymous, 1 year ago

Q1 State the laws of refraction of light . If the speed of light in vacuum is 3 × 10⁸ m/s , find the speed of light in a medium of absolute refractive index 1.5.

Q2 A concave lens has focal length of 20cm . At what distance from the lens should a 5cm tall object be placed so that it forms an image at 15cm from the lens ? Also , calculate the size of the image formed.

Q3 What is the relationship between the S.I unit of power of the lens and S.I unit of focal length ?

Q4 the absolute refractive index of benzene and kerosene are 1.50 and 1.44 respectively. What is the refractive index of benzene with respect to kerosene.

Q5 Define magnification . Explain why magnification is positive for virtual image and negative for real image ??

Q6 Find the position of an object , which when placed in front of a concave mirror of focal length 10cm produces a virtual image , which is twice the size of the object .

Last question
Q7 For an object placed at a distance of 20cm from the pole of the mirror , an image is formed , 40cm farther from the object on the same side.

a)What is the nature of the mirror ? Give reason for your answer.
b)Is the image real or virtual ? Give reason for your answer .
c)Draw a ray diagram to show the image formed.
d)Calculate the focal length of the mirror used.

Please answer all the questions . Please. Thank you in advance.


kvnmurty: you need to write one query - in one questi on of brainly.. writing so many in one question is difficult for the writers and for the moderators... please understand.

Answers

Answered by kvnmurty
18
Laws of refraction

1.  Sin i / Sin r = refractive index of the medium.
 2.  Light rays bend when they are incident on a boundary between two media.
3.  The incident ray, refracted ray, and the normal to the boundary at the point of incidence, are all in one plane.

Refractive index = 1.5 = speed of light in vacuum / speed of light in glass
         => speed of light in glass = 3 * 10^8 /1.5 = 2 * 10^8 m/sec
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 concave lens.  -  forms virtual small erect images

        h = 5 cm      u = ?    v = -15 cm     f = - 20  cm
     1/v - 1/u = 1/f          =>  1/u = 1/v - 1/f = -1/15 +1/20 = -1/60
           u = -60 cm       
             m = h'/h = v/u = -15/-60 = 1/4
               h' = 5/4 cm
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     SI unit of Power of lens = a Dioptre  = D
     SI unit of focal length =  meter.
         Power =  1 / focal length
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Relative refractive index of medium 2 wrt medium 1 = μ₂₁
                   μ₂₁ = μ₂ / μ₁      = 1.50 / 1.44 = 1.0417
===========================================

magnification for lenses  (magnification for mirrors is defined differently)
       m = h'/h = v/u
  A virtual image is always erect and in the same direction as the object.  Hence h and h' have the same sign.  Also,  the image and the object are on the same side.  Hence the signs of u and v are same.

A real image is inverted on the other side of lens.  Hence,  h and h' are in opposite directions.  Also, u and v are on opposite sides of lens.  Hence m is negative.
=============================================
   u = ?   f = -10 cm  for concave mirror,    m = -v/u = -h/h
                   v = -2 u 
           1/f = 1/v + 1/u          =>  -1/10 = -1/2u +1/u = 1/2u
                  u = -5 cm
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 u = -20 cm    v = -(20 + 40) = -60 cm

d)         1/f = -1/20 - 1/60 = -1/15         => f = -15 cm

a)  mirror is concave mirror as the focal length is negative.  Convex mirror always forms images on the other side (behind) of mirror.  ie., v is always positive.

b)  m = -(-60)/(-20) = -3 =  h'/h   image is inverted  as   h' = -3 h
     concave mirror forms erect images when object is within focal length. otherwise, inverted images are formed.

c)    see diagram enclosed.


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