$A$ beaker contains water up to a height of $h_1$ and kerosene above the water up to a height of $h_2$. The total height is $(h_1 + h_2)$. The refractive index of water is $\mu_1$ and that of kerosene is $\mu_2$. What is the apparent shift of the bottom of the beaker when viewed from above?

  • A
    $\left( {1 - \frac{1}{{{\mu _1}}}} \right)\,{h_2} + \left( {1 - \frac{1}{{{\mu _2}}}} \right)\,{h_1}$
  • B
    $\left( {1 + \frac{1}{{{\mu _1}}}} \right)\,{h_1} - \left( {1 + \frac{1}{{{\mu _2}}}} \right)\,{h_2}$
  • C
    $\left( {1 - \frac{1}{{{\mu _1}}}} \right)\,{h_1} + \left( {1 - \frac{1}{{{\mu _2}}}} \right)\,{h_2}$
  • D
    $\left( {1 + \frac{1}{{{\mu _1}}}} \right)\,{h_2} - \left( {1 + \frac{1}{{{\mu _2}}}} \right)\,{h_1}$

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$A$ bubble in a glass slab $(\mu = 1.5)$ when viewed from one side appears at $5 \ cm$ and from the other side appears at $2 \ cm$. The thickness of the slab is .... $cm$.

Two parallel rays of red and violet colour pass through a glass slab. Which of the following is correct?

Two light rays initially in the same phase travel through two media of equal length $L$ having refractive indices $\mu_1$ and $\mu_2$ (where $\mu_1 > \mu_2$) as shown in the figure. If the wavelength of the light rays in air is $\lambda$,the phase difference of the emerging rays is given by:

$A$ convex lens is placed $10\, cm$ from a light source and it forms a sharp image on a screen,kept $10\, cm$ from the lens. Now,a glass block (refractive index $\mu = 1.5$) of $1.5\, cm$ thickness is placed in contact with the light source. To get the sharp image again,the screen is shifted by a distance $d$. Then $d$ is:

Time taken by the sunlight to pass through a slab of $4 \ cm$ and refractive index $1.5$ is . . . . . . $s$.

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