Two identical thin Plano-convex glass lenses (refractive index 1.5 ) each having radius of curvature of 20 cm are placed with their convex surfaces in contact at the center. The intervening space is filled with oil of

Correct answer: C. - 50cm Len's maker's formula (1)/(f)=(μ - 1)( 1 R _ 1 - 1 R _ 2 ) 1 f _ 1 = (μ - 1)(1)/(20) ⇒ (1.5 - 1)(1)/(20) = (1)/(40) 1 f _ 1 = (1)/(40) For lens 1 f _ 1 + 1 f _ 2 = (1)/(40) + (1)/(40) = (1)/(20) For concave 1 f _ 3 = (1.7 - 1) ( - (1)/(R) - (1)/(R) ) = (- 0.7 × 2)/(R) 1 f _ eq = (1)/(20) + ((- 0.7 × 2)/(R)) = (1)/(20) - (1.4)/(20) f _ eq = - (20)/(0.4) = - 50cm

NEET 2015

Exam: NEET
Year: 2015
Subject: Physics
Chapter: Ray Optics

NEET 2015 · Physics · Ray Optics

Two identical thin Plano-convex glass lenses (refractive index 1.5...

Two identical thin Plano-convex glass lenses (refractive index

\text{1}\text{.5}

) each having radius of curvature of

\text{20 cm}

are placed with their convex surfaces in contact at the center. The intervening space is filled with oil of refractive index

\text{1}\text{.7}

The focal length of the combination is $Two identical thin Plano-convex glass lenses (refractive index 1.5 ) each having radius of curvature of 20 cm are$

Options

A $- 20cm$
B $- 25cm$
C $- 50cm$
D $50cm$

Correct answer

C. $- 50cm$

Step-by-step solution

$Two identical thin Plano-convex glass lenses (refractive index 1.5 ) each having radius of curvature of 20 cm are$
Len's maker's formula

\dfrac{1}{f}=\left(\mu - 1\right)\left(\dfrac{1}{{R}_{1}} - \dfrac{1}{{R}_{2}}\right)

\dfrac{1}{{f}_{1}} = \left(\mu - 1\right)\dfrac{1}{20} ⇒ \left(1.5 - 1\right)\dfrac{1}{20} = \dfrac{1}{40}

\dfrac{1}{{f}_{1}} = \dfrac{1}{40}

For lens

\dfrac{1}{{f}_{1}} + \dfrac{1}{{f}_{2}} = \dfrac{1}{40} + \dfrac{1}{40} = \dfrac{1}{20}

For concave

\dfrac{1}{{f}_{3}} = \left(1.7 - 1\right) ( - \dfrac{1}{R} - \dfrac{1}{R} ) = \dfrac{- 0.7 \times 2}{R}

\dfrac{1}{{f}_{eq}} = \dfrac{1}{20} + \left(\dfrac{- 0.7 \times 2}{R}\right)

= \dfrac{1}{20} - \dfrac{1.4}{20}

{f}_{eq} = - \dfrac{20}{0.4} = - 50cm

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