A thin convergent glass lens μ =1.5 has a power of +5.0 D . When this lens is immersed in a liquid of refractive index μ 1 it acts as a diverging lens of focal length 100 cm . The value of μ l should be

Question

A thin convergent glass lens μ =1.5 has a power of +5.0 D . When this lens is immersed in a liquid of refractive index μ 1 it acts as a diverging lens of focal length 100 cm . The value of μ l should be (A)  (3/2)  (B)  (4/3)  (C)  (5/3)  (D)  2

Tardigrade Admin · Accepted Answer

When the lens in air, we have Pa =(1/fa)=(μg-μa/μa)[(1/R1)-(1/R2)] When the lens is in liquid, we have P1 =(1/fl)=(μg-μl/μl)[(1/R1)-(1/R2)] Given, Pa = 5, P1 = - 1, μa = 1, μg = 1.5 On soiving, we get, μl = (5/3)

Q. A thin convergent glass lens $μ = 1.5$ has a power of $+ 5.0 D$ . When this lens is immersed in a liquid of refractive index $μ_{1}$ it acts as a diverging lens of focal length $100 c m$ . The value of $μ_{l}$ should be

$\frac{3}{2}$

$\frac{4}{3}$

$\frac{5}{3}$

$2$

Q. A thin convergent glass lens μ=1.5 has a power of +5.0D . When this lens is immersed in a liquid of refractive index μ1​ it acts as a diverging lens of focal length 100cm . The value of μl​ should be

23​

34​

35​

2

When the lens in air, we have Pa​=fa​1​=μa​μg​−μa​​[R1​1​−R2​1​] When the lens is in liquid, we have P1​=fl​1​=μl​μg​−μl​​[R1​1​−R2​1​] Given, Pa​=5,P1​=−1, μa​=1,μg​=1.5 On soiving, we get, μl​=35​

Q. A thin convergent glass lens $μ = 1.5$ has a power of $+ 5.0 D$ . When this lens is immersed in a liquid of refractive index $μ_{1}$ it acts as a diverging lens of focal length $100 c m$ . The value of $μ_{l}$ should be

$\frac{3}{2}$

$\frac{4}{3}$

$\frac{5}{3}$

$2$