Question

A point object is kept 10 cm away from one of the surfaces of a thick double convex lens of refractive index 1.5 and radii of curvature 10 cm and 8 cm . Determine Focal length.​

Answer 1

Solution :

As per the given data ,

• Image distance ( u) = - 10 cm
• Refractive index of the lens (μ) = 1.5
• R 1 = 10 cm
• R 2 = - 8 cm

Note : Distances are always measured from the optical center of the lens .

By using the lens makers formula ,

$\leadsto\mathtt{\dfrac{1}{f} = (n-1) \bigg[ \dfrac{1}{R_1} - \dfrac{1}{R_2}\bigg] }$

$\leadsto\mathtt{\dfrac{1}{f} = (1.5-1) \bigg[ \dfrac{1}{10} + \dfrac{1}{8}\bigg]}$

$\leadsto\mathtt{\dfrac{1}{f} = 0.5 \bigg[ \dfrac{18}{80}\bigg]}$

$\leadsto\mathtt{\dfrac{1}{f} = 0.5 \times 0.225 }$

$\leadsto\mathtt{f = \dfrac{1}{0.1125} }$

$\leadsto\mathtt{f = 8.8 cm }$

The focal length of the object is 8.8 cm .

Additional information :

• The focal length of a convex lens is positive
• The focal length of a concave lens is negative

Lens formula,

• 1 / f = 1 / v - 1/ u

Magnification ,

• m = v / u