A liquid of refractive index 1-6 is contained in the cavity of a glass specimen of refractive index 1-5 as shown in Fig- If each of the curved surfaces has a radius of curvature of 0-20 m- the arrangement behaves as aConverging lens of focal length 0-25 mDiverging lens of focal length 0-25 mDiverging lens of focal length 0-17 mConverging lens of focal length 0-72 m

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Toppr Admin · Accepted Answer

Power of liquid lens-1-6-x2212-1-20-20-610-xD7-10-6-xA0-DPower of concave lens-x2212-1-5-x2212-1-x2212-0-5-xD7-10-xA0-DTotal power of two concave lenses -x2212-10-xA0-DPower of system-x2212-10-xA0-D-6-xA0-D-x2212-4-xA0-DForce length -1-x2212-4-0-25-xA0-m-

Power of liquid lens=(1.6−1)(20.20)=610×10=6 DPower of concave lens=−(1.5−1)=−0.5×10 DTotal power of two concave lenses =−10 DPower of system=−10 D+6 D=−4 DForce length =1−4=0.25 m.

Power of liquid lens
$= (1.6 - 1) (\frac{2}{0.20}) = \frac{6}{10} \times 10 = 6 D$
Power of concave lens
$= - (1.5 - 1) = - 0.5 \times 10 D$
Total power of two concave lenses $= - 10 D$
Power of system
$= - 10 D + 6 D = - 4 D$
Force length $= \frac{1}{- 4} = 0.25 m$ .