# Question is in the pic below and the answer is 12 cm. Can you please try it?

## Mar 11, 2017

(4) $12 c m$

#### Explanation:

As shown in the figure when a ray of light, parallel to the principal axis hits the first glass piece, it acts as a thin slab of glass. As the ray enters the slab, it moves from rarer to denser (air to glass) medium and therefore is refracted and bends towards the normal in the denser medium. As it emerges on the other side, it moves from a denser to rarer medium and bends away from the normal in the rarer medium. Over all straight line path of ray is un-deviated. Ignoring the thickness of glass the minor lateral shift in the path of ray of light is ignored. Consequently, it continues its travel parallel to the principal axis and hits the concave mirror, gets reflected and passes through the focus of the mirror.
Focal length of the mirror is given as $f = - 20 c m$ .....(1)

We also know that focal length of concave mirror is related to radius of curvature of mirror $r$ as
$f = \frac{- r}{2}$
$\therefore r = 40 c m$
this is keeping in view the sign convention.

When air between the glasses is replaced by water, it acts as a convex lens of water $\mu = \frac{4}{3}$ with one surface polished to act as a concave mirror.

Using the lens maker formula
$\frac{1}{f} = \left(\mu - 1\right) \left(\frac{1}{R} _ 1 - \frac{1}{R} _ 2\right)$

Recalling that both glass pieces are similar, also keeping in view the sign convention we get
$\frac{1}{{f}_{\text{lens}}} = \left(\frac{4}{3} - 1\right) \left(\frac{1}{40} + \frac{1}{40}\right)$
$\implies \frac{1}{{f}_{\text{lens}}} = \frac{1}{3} \times \frac{1}{20} = \frac{1}{60}$ ...(2)

The parallel ray of light hits water lens gets refracted, hits mirror, is reflected and gets refracted again as it travels in the reverse direction in the water lens.
If $\frac{1}{F}$ is the effective focal length of the combination we have
$\frac{1}{F} = 2 \times \frac{1}{{f}_{\text{lens}}} + \frac{1}{f}$
From (1) and (2) Keeping in view the sign convention
$\implies \frac{1}{F} = 2 \times \frac{1}{60} + \frac{1}{20}$
$\implies \frac{1}{F} = \frac{1}{30} + \frac{1}{20} = \frac{2 + 3}{60} = \frac{1}{12}$
$F = 12 c m$
All rays of light from infinity, parallel to principal axis will form an image at $12 c m$

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Alternative steps after equation (2)

For lens the image is formed at a distance of $60 c m$ on the right.
This acts as object for the concave mirror. Using the mirror equation and keeping in view the sign convention
$\frac{1}{f} + \frac{1}{O} = \frac{1}{I}$
$\implies \frac{1}{20} + \frac{1}{60} = \frac{1}{I} _ m$
$\implies \frac{1}{I} _ m = \frac{3 + 1}{60} = \frac{1}{15}$

Image is formed at a distance of $15 c m$
This acts as object for the convex lens again. The final image is formed as
$\frac{1}{f} = \frac{1}{O} + \frac{1}{I}$
$\frac{1}{60} = \frac{1}{- 15} + \frac{1}{I} _ f$
$\frac{1}{I} _ f = \frac{1}{60} + \frac{1}{15} = \frac{1 + 4}{60} = \frac{1}{12}$

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Cartesian Sign Convention

1. All figures are drawn with light traveling from left to right.
2. All distances are measured from a reference surface. Distances to the left of the surface are negative.
3. The refractive power of a surface that makes light rays more convergent is positive. The focal length of such a surface is positive.
4. The distance of a real object is negative.
5. The distance of a real image is positive.
6. Heights above the optic axis are positive.