$A$ cubic block of glass with refractive index $n_1$ is in contact with the surface of water with refractive index $n_2$. $A$ ray of light is incident on the vertical face of the block (see figure). After refraction,total internal reflection at the base,and refraction at the opposite vertical face,the ray emerges at an angle $\theta$. The value of $\theta$ is given by:

  • A
    $\sin\, \theta < \sqrt {{n_1}^2 - {n_2}^2} $
  • B
    $\tan\, \theta < \sqrt {{n_1}^2 - {n_2}^2} $
  • C
    $\sin\, \theta < \frac{1}{{\sqrt {{n_1}^2 - {n_2}^2} }}$
  • D
    $\tan\, \theta < \frac{1}{{\sqrt {{n_1}^2 - {n_2}^2} }}$

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Assertion : $A$ concave mirror and convex lens both have the same focal length in air. When they are submerged in water,they will have same focal length.
Reason : The refractive index of water is smaller than the refractive index of air.

$A$ plano-convex lens is made of a material of refractive index $n$. When a small object is placed $30 \ cm$ away in front of the curved surface of the lens,an image of double the size of the object is produced. Due to reflection from the curved surface of the lens,another faint image is observed at a distance of $10 \ cm$ away from the lens. Which of the following statement$(s)$ is(are) true?
$(A)$ The refractive index of the lens is $2.5$
$(B)$ The radius of curvature of the convex surface is $45 \ cm$
$(C)$ The faint image is erect and real
$(D)$ The focal length of the lens is $20 \ cm$

$A$ small piece of wire bent into an $L$ shape with upright and horizontal portions of equal lengths,is placed with the horizontal portion along the axis of a concave mirror whose radius of curvature is $10 \, cm$. If the bend is $20 \, cm$ from the pole of the mirror,then the ratio of the lengths of the images of the upright and horizontal portions of the wire is:

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$A$ convex lens of focal length $30\,cm$,a concave lens of focal length $120\,cm$,and a plane mirror are arranged as shown. For an object kept at a distance of $60\,cm$ from the convex lens,the final image,formed by the combination,is a real image at a distance of:

$A$ ray of light moving along the vector $\vec{v} = -\hat{i} - 2\hat{j}$ undergoes refraction at an interface of two media,which is the $x-z$ plane. The refractive index for $y > 0$ is $\mu_1 = 2$,while for $y < 0$,it is $\mu_2 = \sqrt{5}/2$. Find the unit vector along which the refracted ray moves.

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