A convex lens of refractive index 32 has a power of 2.5 D in air. If it is placed in a liquid of refractive index 2, then the new power of the lens is
A
1.5 D
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
1.2 D
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
−1.25 D
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
1.25 D
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is C−1.25 D Let the radius of curvature of the convex lens is R1 and R2
Focal length of a convex lens having power 2.5 D,=12.5 m
Focal length of a lens in air is given by, 1f=(μ−1)(1R1−1R2) ⇒2.5=1f=(32−1)(1R1−1R2).....(i)
Focal length of the lens in a liquid of refractive index 2 is ⇒1f′=⎛⎜
⎜
⎜⎝322−1⎞⎟
⎟
⎟⎠(1R1−1R2)....(ii)
Dividing the equation (i) by equation (ii), 2.5f′=0.5−0.25 ⇒1f′=−2.5×0.250.5=−1.25 D