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A convex lens of focal length 0.24 m and...

A convex lens of focal length 0.24 m and of refractive index 1.5 is completely immersed in water of refractive 1.33. Find the changes in the focal length of the lens.

Text Solution

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Lens in air, `(1)/(f_(g)) \= ( n _(g)-1) ((1)/(R_(1))-(1)/(R_(2)) )`, `(f_(g) = `focal length in air )
`(1)/(0.24) = ( 1.5-1) ((1)/(R_(1))-(1)/(R_(2)))`
Lens in water, ` (1)/( f_(gw)) = ( (n_(g))/(n_(w))-1) = ( (1)/( R_(1)) - ( 1)/( R_(2)))`, `( f_(gw)` = focal length of lens in water )
`(1)/( f_(gw)) = ((1.5)/(1.3) -1) ( (1 )/( R_(1)) - ( 1)/( R_(2))) ` ....(ii)
Dividing equation (1) by (2),
`f_(gw) = 0.939m`
`:.` Change in focal length `= 0.699m ( with unit )
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Knowledge Check

  • A double convex lens of focal length 20 cm is made of glass of refractive index 3/2. When placed completely in water (._(a)mu_(w)=4//3) , its focal length will be :

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    C
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