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Figure 9.30 shows 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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1. Given Data
- Refractive Index of Glass (ng): 1.50
- Focal Length of Combination (F): 45.0 cm (with liquid)
- Focal Length of Convex Lens alone (f1): 30.0 cm (without liquid)
2. Calculating Radius of Curvature (R)
First, we find the radius of curvature of the equiconvex lens using the Lens Maker's Formula.
1/f1 = (ng - 1) × (2/R)
Substituting the values:
- 1/30 = (1.50 - 1) × (2/R)
- 1/30 = 0.5 × (2/R)
- 1/30 = 1/R
- R = 30 cm
3. Calculating Focal Length of Liquid Lens (f2)
The system behaves as a combination of two lenses in contact: the glass lens (f1) and the liquid lens (f2). The combined focal length is F.
1/F = 1/f1 + 1/f2
Substituting the values:
- 1/45 = 1/30 + 1/f2
- 1/f2 = 1/45 - 1/30
- 1/f2 = (2 - 3) / 90
- 1/f2 = -1 / 90
- f2 = -90 cm
4. Calculating Refractive Index of Liquid (nl)
The liquid forms a plano-concave lens. One surface matches the curvature of the convex lens (R1 = -R = -30 cm) and the other is flat (R2 = ∞).
Using the Lens Maker's Formula for the liquid lens:
1/f2 = (nl - 1) × (1/R1 - 1/R2)
Substituting the values:
- -1/90 = (nl - 1) × (1/-30 - 1/∞)
- -1/90 = (nl - 1) × (-1/30)
Dividing both sides by (-1/30):
- 30 / 90 = nl - 1
- 1/3 = nl - 1
- nl = 1 + 0.333
- nl ≈ 1.33
Refractive Index of Liquid = 1.33
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