$A$ fish in an aquarium,$30 \, cm$ deep in water,can see a light bulb kept $50 \, cm$ above the surface of water. The fish can also see the image of this bulb in the reflecting bottom surface of the aquarium. The total depth of water is $60 \, cm$. Then the apparent distance between the two images seen by the fish is $(\mu_w = 4/3)$.

  • A
    $140 \, cm$
  • B
    $\frac{760}{3} \, cm$
  • C
    $\frac{280}{3} \, cm$
  • D
    $\frac{380}{3} \, cm$

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$List-I$$List-II$
$(P)$ If $n=2$ and $\alpha=180^{\circ}$,then all the possible values of $\theta_0$ will be$(1)$ $30^{\circ}$ and $0^{\circ}$
$(Q)$ If $n=\sqrt{3}$ and $\alpha=180^{\circ}$,then all the possible values of $\theta_0$ will be$(2)$ $60^{\circ}$ and $0^{\circ}$
$(R)$ If $n=\sqrt{3}$ and $\alpha=180^{\circ}$,then all the possible values of $\phi_0$ will be$(3)$ $45^{\circ}$ and $0^{\circ}$
$(S)$ If $n=\sqrt{2}$ and $\theta_0=45^{\circ}$,then all the possible values of $\alpha$ will be$(4)$ $150^{\circ}$
$(5)$ $0^{\circ}$

If the polarising angle of a piece of glass for green light is $54.74^o$,then the angle of minimum deviation for an equilateral prism made of the same glass is......$^o$ (Given $\tan 54.74^o = 1.414$)

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