$A$ tank is filled with water to a height of $16 \text{ cm}$. Find the apparent depth of a needle lying at the bottom of the tank as measured by a microscope. The refractive index of water $(\mu_{w})$ is $\frac{4}{3}$. (in $\text{ cm}$)

  • A
    $8.0$
  • B
    $10.6$
  • C
    $12.0$
  • D
    $9.4$

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The refractive indices of glass and water with respect to air are $3/2$ and $4/3$ respectively. The refractive index of glass with respect to water is:

Consider slabs of three media $A, B$ and $C$ arranged as shown in the figure. The refractive index $(R.I.)$ of $A$ is $1.5$ and that of $C$ is $1.4$. If the number of waves in $A$ is equal to the number of waves in the combination of $B$ and $C$,then the refractive index of $B$ is:

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An electromagnetic wave,whose wave normal makes an angle of $45^{\circ}$ with the vertical,travelling in air strikes a horizontal liquid surface. While travelling through the liquid it gets deviated through $15^{\circ}$. What is the speed of the electromagnetic wave in the liquid,if the speed of electromagnetic wave in air is $3 \times 10^8 \ m/s$? $(\sin 30^{\circ} = 0.5, \sin 45^{\circ} = \frac{1}{\sqrt{2}})$

If light passes near a massive object,the gravitational interaction causes a bending of the ray. This can be thought of as happening due to a change in the effective refractive index of the medium given by $n = 1 + \frac{2GM}{rc^2}$,where $r$ is the distance of the point of consideration from the centre of the mass of the massive body,$G$ is the universal gravitational constant,$M$ is the mass of the body,and $c$ is the speed of light in vacuum. Considering a spherical object,find the deviation of the ray from the original path as it grazes the object.

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Two light beams fall on a transparent material block at points $1$ and $2$ with angles $\theta_1$ and $\theta_2$ respectively,as shown in the figure. After refraction,the beams intersect at point $3$,which is exactly on the interface at the other end of the block. Given: the distance between $1$ and $2$ is $d = 4\sqrt{3} \text{ cm}$ and $\theta_1 = \theta_2 = \cos^{-1}\left(\frac{n_2}{2n_1}\right)$,where $n_2$ is the refractive index of the block and $n_1$ is the refractive index of the outside medium $(n_2 > n_1)$. Find the thickness of the block in $\text{cm}$.

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