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Geometrical Optics

Section: Physics

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Question : 44 of 56

Marks: +1, -0

A biconvex lens is formed with two planoconvex lenses as shown in the figure. Refractive index n of the first lens is 1.5 and that of the second lens is 1.2. Both curved surface are of the same radius of curvature R = 14 cm. For this biconvex lens, foran object distance of 40 cm, the image distance will be

[JEE Adv 2012 P1]

–280.0 cm
40.0 cm
21.5 cm
13.3 cm

Solution:

For this biconvex lens, for an object distance of 40 cm, the image distance is calculated as follows:

P_T

=

(1.5-1)\left(\frac{1}{14}-0\right)

+

(1.2-1)\left(0-\frac{1}{-14}\right)

=

\frac{0.5}{14}+\frac{0.2}{14}

=

\frac{1}{20}

That is, f = +20 cm. Therefore,

\frac{1}{v} - \frac{1}{-40}

=

\frac{1}{20}

⇒

\frac{1}{v}

=

\frac{1}{20}-\frac{1}{40}

=

\frac{1}{40}

Therefore, v = 40 cm.

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