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A thin equiconvex spherical glass lens (µ = 3/2) of radius of curvature 30 cm is placed on the x-axis with its optical centre at x = 40 cm and principal axis coinciding with the x-axis. A light ray given by the equation 39y = – x + 1 (x and y are in cm) is incident on the lens, in the direction of positive x-axis. The correct alternative(s) isA) equation of refracted ray is 39y=x-1B) equation of refracted ray is 130y=x-170C) equation of the refracted ray if space on the right side is filled with a liquid of refractive index 4/3 390y+x+350=0D)equation of the refracted ray if space on the right side is filled with a liquid of refractive index 4/3 390y-x+350=0

A thin equiconvex spherical glass lens (µ = 3/2) of radius of curvature 30 cm is placed on the x-axis with its optical centre at x = 40 cm and principal axis coinciding with the x-axis. A light ray given by the equation 39y = – x + 1 (x and y are in cm) is incident on the lens, in the direction of positive x-axis. The correct alternative(s) isA) equation of refracted ray is 39y=x-1B) equation of refracted ray is 130y=x-170C) equation of the refracted ray if space on the right side is filled with a liquid of refractive index 4/3 390y+x+350=0D)equation of the refracted ray if space on the right side is filled with a liquid of refractive index 4/3 390y-x+350=0

Grade:12

1 Answers

Arun
25750 Points
3 years ago
When air is on both sides of the lens,
20
1
​ 
 − 
−20
1
​ 
 = 
f
1
​ 
 
Hence,  
f
1
​ 
 = 
10
1
​ 
 =(μ 
lens
​ 
 −1)( 
R
1
​ 
 − 
−R
1
​ 
 )
Hence, R=10cm
2. When liquid is inside the vessel.
For the refraction from bottom side of the lens, 
v
μ 
g
​ 
 
​ 
 − 
u
μ 
1
​ 
 
​ 
 = 
R
μ 
g
​ 
 −μ 
1
​ 
 
​ 
 
hence  
v
1.5
​ 
 = 
20
3−3μ 
1
​ 
 
​ 
 
Now this v acts as an object for the refraction from top surface of the lens.
Thus 
1
​ 
 
1
​ 
 − 
v
1.5
​ 
 = 
−10
1−1.5
​ 
 
Final image is formed 30cm from lens, 
Thus v 
1
​ 
 =30cm
Hence, μ 
1
​ 
 =1.11

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