The refractive index of a material of a planoconcave lens is (5/3) , the radius of curvature is 0.3 m . The focal length of the lens in air is

Question

The refractive index of a material of a planoconcave lens is (5/3) , the radius of curvature is 0.3 m . The focal length of the lens in air is (A) -0.45m (B) -0.6m (C) -0.75m (D) -1.0m

Tardigrade Admin · Accepted Answer

Lens maker's formula (1/f)=(μ - 1) [(1/R1) - (1/R2)] Where , R2= ∈ fty, R1=0.3 m ∴ (1/f)=((5/3) - 1)((1/0.3) - (1/∈ fty)) ⇒ (1/f)=(2/3)× (1/0.3) Or f=0.45 m

Q. The refractive index of a material of a planoconcave lens is $\frac{5}{3}$ , the radius of curvature is $0.3 m$ . The focal length of the lens in air is

$- 0.45 m$

$- 0.6 m$

$- 0.75 m$

$- 1.0 m$

Lens maker's formula
$\frac{1}{f} = (μ - 1) [\frac{1}{R _{1}} - \frac{1}{R _{2}}]$
Where , $R_{2} = \in f t y, R_{1} = 0.3 m$
$∴ \frac{1}{f} = (\frac{5}{3} - 1) (\frac{1}{0.3} - \frac{1}{\in f t y})$
$\Rightarrow \frac{1}{f} = \frac{2}{3} \times \frac{1}{0.3}$
$O r f = 0.45 m$

Q. The refractive index of a material of a planoconcave lens is 35​ , the radius of curvature is 0.3m . The focal length of the lens in air is

−0.45m

−0.6m

−0.75m

−1.0m