A thin lens of refractive index 1.5 has a focal length 15 cm in air. When the lens is placed in a medium of refractive index 4/3, what is the new focal length?
60 cm
120 cm
70 cm
50 cm
A
60 cm
B
50 cm
C
120 cm
D
70 cm
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Solution
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For the lens in air
1fa=(aμg−1)[1R1−1R2]
⇒115=(1.5−1)[1R1−1R2]
⇒215=1R1−1R2
When the lens is immersed in water
1fw=(aμgaμw−1)[1R1−1R2]
⇒1fw=⎛⎜
⎜
⎜⎝1.543−1⎞⎟
⎟
⎟⎠[1R1−1R2]
⇒1fw[4.54−1][1R1−1R2]
⇒1fw=18×215
⇒1fw=160
⇒fw=60cm.
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