An equi-convex lens with radii of curvature of magnitude r each, is put over a liquid layer poured on top of a plane mirror. A small needle with its tip on the principal axis of the lens is moved along the axis until its inverted real image coincides with the needle itself. The distance of needle from lens is measured to be a. On removing the liquid layer and repeating the experiment, the distance is found to b. Given that two values of distances measured represent the real length values in the two cases, obtain a formula for refractive index of the liquid.

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Here, combined focal length of glass lens and liquid lens, F = a, and Focal length of convex lens, f₁ = b.
If f₂ is focal length of liquid lens, then


The liquid lens is plano-concave lens for which
                           
From               
                    
          
                       

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Angle of prism, A = 60^o 
Refractive index of prism = 1.5 
Given, angle of incidence , i₁=i₂ = $\frac{3}{4}A$ = $\frac{3}{4}{45}^o$ = 45^o
As,                    A + $\mathrm{\delta }$ = i₁ + i₂ 
$\Rightarrow$                60^o + $\mathrm{\delta }$ = 45^o + 45^o 

Then, angle of minimum deviation, $\mathrm{\delta }$ =  90^o - 60^o                                                           = 30^o 
We know that, velocity of light in vacuum = 3$\times$10⁸ m/s 
Now, using the formula, 

                $\mathrm{\mu } = \frac{\sin (\mathrm{A}+{\mathrm{\delta }}_{\mathrm{m}})/2}{\sin \mathrm{A}/2} = \frac{\mathrm{c}}{\mathrm{v}}$
That is,  
                $$\begin{gathered} \mathrm{v} = \mathrm{c} \frac{\sin \mathrm{A}/2}{\sin (\mathrm{A}+{\mathrm{\delta }}_{\mathrm{m}})/2} \\ = 3\times {10}^8 \frac{\sin {30}^{\mathrm{o}}}{\sin ({60}^{\mathrm{o}} + {30}^{\mathrm{o}})/2} \\ = 3\times {10}^8 \frac{\sin {30}^{\mathrm{o}}}{\sin {45}^{\mathrm{o}}} \\ = \frac{ 3\times {10}^8 \times 0.5000}{0.7071} \\ = 2.12 \times {10}^8 \mathrm{m}/\mathrm{s} \end{gathered}$$ 
 
is the required speed of light in the prism.

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For the refraction at convex lens, we have


u = –12 cm; v = ? f = + 10 cm Using lens formula, we have
                                      
or,                                  

Thus, in the absence of the convex mirror, convex lens will form the image I₁, at a distance of 60 cm behind the lens. As the mirror is at a distance of 10 cm from the lens, I₁ will be at a distance of (60 – 10) = 50 cm from the mirror, i.e., MI₁ = 50 cm.
Now, as the final image I₂ is formed at the object itself, the rays after reflection from the mirror retraces its path, i.e., the rays on the mirror are incident normally, i.e., I₁ is the centre of the mirror so that
                                      
and                                

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Let f be the focal length of the equivalent spherical mirror.

We have, 
                          

or,                

Here,            
                   

              

or,                     
The problem is reduced to a simple case where a point object is placed in front of a concave mirror.
Now, using mirror formula
                        we have                     
                   
                       
The image is erect and virtual and 3 lines of object.

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The focal length of the lens having radii of its curvature R₁ and R₂ and refractive index of its material with respect to air μ, f is given by
                     ...(i)
When the lens is put in a transparent medium, the refractive index of the material of the lens with respect to the medium 
Thus focal length of the lens in the medium f' is given by
                                                 ...(ii)
From Eqns. (i) and (ii),
                             

or                                                                 ...(iii)
Thus, change in focal length
 

         

         

          

           

                              ...(iv)
From Equ (iii),
                              
Putting