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An equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0cm. What is the refractive index of the liquid?
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An equiconvex lens (of refractive index 1.50) in contact with a liquid...
Problem:

An equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0cm. What is the refractive index of the liquid? Explain in details.


Solution:


Step 1: Understand the Problem

The problem deals with the measurement of the refractive index of a liquid. The refractive index of the lens is given as 1.50. The experiment involves finding the position of the inverted image of a needle placed on the principal axis of the lens. The needle is first placed at a distance of 45.0cm from the lens when the liquid is present and then at a distance of 30.0cm when the liquid is removed. The refractive index of the liquid needs to be calculated.


Step 2: Identify the Relevant Formula

The relevant formula for this problem is the lens formula:

1/f = 1/v - 1/u

where f is the focal length of the lens, v is the distance of the image from the lens, and u is the distance of the object from the lens.


Step 3: Apply the Formula

Let the refractive index of the liquid be n and the radius of curvature of the lens be R. When the needle is placed at a distance of 45.0cm from the lens, we can write:

1.50 = (n-1) (1/R - 1/(-R))

1.50 = (n-1) (2/R)

1/R = (n-1) (2/1.50R)

R = 2nR/3 - 2R/3

u = -45cm, v = -45cm

1/f = (n-1) (1/R)

1/f = (n-1) (3/2nR)

f = 3R/2(n-1)

1/v - 1/u = 1/f

1/v - 1/-45 = 2(n-1)/3nR

v = -90nR/(3n-2)

When the liquid is removed and the needle is placed at a distance of 30.0cm from the lens, we can write:

1.50 = (1/R)

R = 2/3 cm

u = -30cm, v = -30cm

1/f = (1/R)

f = R

1
Community Answer
An equiconvex lens (of refractive index 1.50) in contact with a liquid...
Figure shows an equiconvex lens.png
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Focal length of the convex lens, f1 = 30 cm 
The liquid acts as a mirror. Focal length of the liquid = f2 
Focal length of the system (convex lens + liquid), f = 45 cm 
For a pair of optical systems placed in contact, the equivalent focal length is given as:
Let the refractive index of the lens be u, and the radius of curvature of one surface be R. Hence, the radius of curvature of the other surface is −R.
Focal length of the convex lens, f1 = 30 cm 
The liquid acts as a mirror. Focal length of the liquid = f2 
Focal length of the system (convex lens + liquid), f = 45 cm 
For a pair of optical systems placed in contact, the equivalent focal length is given as:
Let the refractive index of the lens be u, and the radius of curvature of one surface be R. Hence, the radius of curvature of the other surface is −R.
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An equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0cm. What is the refractive index of the liquid?
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An equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0cm. What is the refractive index of the liquid? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about An equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0cm. What is the refractive index of the liquid? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0cm. What is the refractive index of the liquid?.
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