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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

```
3 years ago

Arun
25768 Points
```							When air is on both sides of the lens,201​  − −201​  = f1​  Hence,  f1​  = 101​  =(μ lens​  −1)( R1​  − −R1​  )Hence, R=10cm2. When liquid is inside the vessel.For the refraction from bottom side of the lens, vμ g​  ​  − uμ 1​  ​  = Rμ g​  −μ 1​  ​  hence  v1.5​  = 203−3μ 1​  ​  Now this v acts as an object for the refraction from top surface of the lens.Thus v 1​  1​  − v1.5​  = −101−1.5​  Final image is formed 30cm from lens, Thus v 1​  =30cmHence, μ 1​  =1.11
```
7 months ago
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