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A double convex thin lens made of glass of refractive index 1.6 has radii of curvature 15 cm each. The focal length of this lens when immersed in a fluid of refractive index 1.63 is​

  • a)
    – 407 cm

  • b)
    125 cm

  • c)
    25 cm

  • d)
    + 250 cm

Correct answer is option 'A'. Can you explain this answer?
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Using the lens maker's formula:

1/f = (n-1) * (1/R1 - 1/R2)

where f is the focal length, n is the refractive index of the lens, and R1 and R2 are the radii of curvature of the two lens surfaces. Since the lens is double convex, both radii of curvature are positive and equal to 15 cm. We can plug in the values:

1/f = (1.6-1) * (1/15 - 1/15)
1/f = 0

This means that the focal length of the lens is infinite when it is in air (or any medium with a lower refractive index than the lens). When the lens is immersed in a fluid with a higher refractive index than the lens (such as the one in this problem), the focal length will be positive and finite. Using the same formula with the new refractive index:

1/f' = (1.63-1.6) * (1/15 - 1/15)
1/f' = 0.002/15
f' = 750 cm

Therefore, the focal length of the lens when immersed in the fluid is 750 cm.
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A double convex thin lens made of glass of refractive index 1.6 has radii of curvature 15 cm each. The focal length of this lens when immersed in a fluid of refractive index 1.63 isa)– 407 cmb)125 cmc)25 cmd)+ 250 cmCorrect answer is option 'A'. Can you explain this answer?
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