Question

Two similar thin equi-convex lenses, of focal length feach, are kept coaxially in contact with each other 16 such that the focal length of the combination is F₁. When the space between the two lenses is filled with glycerin (which has the same refractive index (μ = 1.5) as that ofglass) then the equivalent focal length is F₂. The ratio F₁: F₂ will be :
(1) 2:1
(2) 1:2
(3) 2:3
(4) 3:4

Answer 1

The ratio of focal lengths is $\bold{F_{1}:F_{2}=1:2}$

Step by step explanation:

The equi-convex lenses is a convex lens have two opposite spherical surfaces which have the same shape.

From the given,

The equivalent focal lengths will be $F_{1}\,and\, F_{2}$ in air and glycerin respectively.

$\frac{1}{F_{1}}=\frac{1}{f}+\frac{1}{f}$

   $\Rightarrow \frac{2}{f}$

It behaves like a diverging lens when glycerin is filled inside the lens.

Then focal length will be,

                     $\Rightarrow \frac{1}{F_{2}}=\frac{1}{f}+\frac{1}{f}-\frac{1}{f}$

                     $\Rightarrow \frac{1}{f}$

                    $\frac{F_{1}}{F_{2}}=\frac{\frac{2}{f}}{\frac{1}{f}}=\frac{1}{2}$

Therefore, The ratio of focal lengths is $\bold{F_{1}:F_{2}=1:2}$

Answer 2

Explanation:

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