Two plano-concave lenses (1 and 2) of glass of refractive index 1.5 ha...
Given data:
- Refractive index of glass lens, ng = 1.5
- Radii of curvature of lens 1, R1 = -25 cm (negative sign indicates concave surface)
- Radii of curvature of lens 2, R2 = -20 cm (negative sign indicates concave surface)
- Refractive index of liquid, nl = 4/3
To find: The combined lens is a concave lens of focal length.
Solution:
When two thin lenses are in contact with each other, the effective focal length of the combination can be calculated using the lens formula:
1/f = (ng - nl) * (1/R1 - 1/R2)
- Since the lenses are in contact, their separation is zero, and the space between them is filled with liquid. Therefore, the effective refractive index between the lenses will be that of the liquid, i.e., nl.
- When light travels from a medium with a higher refractive index to a medium with a lower refractive index (ng > nl), the lens behaves as a diverging lens or a concave lens.
Using the lens formula, we can calculate the effective focal length of the combination:
1/f = (1.5 - 4/3) * (1/-25 - 1/-20)
Simplifying the equation:
1/f = (-1/6) * (4/500 - 5/500)
1/f = (-1/6) * (-1/500)
1/f = 1/3000
f = 3000 cm
The focal length of the combination is 3000 cm or 30 m.
Since the combination behaves as a concave lens, the negative sign indicates that the focal length is negative, i.e., the combination is a concave lens of focal length -30 m or -3000 cm.
Therefore, the correct option is (c) concave lens of focal length 66.6 cm.
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