# 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.

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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.

2014-12-23T00:22:38+05:30

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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
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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.
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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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