$A$ laser beam falls on a crystal ball of radius $R$ as shown in the figure. What is its refractive index?

  • A
    $1$
  • B
    $1.5$
  • C
    $1.7$
  • D
    $2$

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Similar Questions

$A$ luminous point object $O$ is placed at a distance $2R$ from the spherical boundary separating two transparent media of refractive indices $n_1$ and $n_2$ as shown,where $R$ is the radius of curvature of the spherical surface. If $n_1 = \frac{4}{3}$,$n_2 = \frac{3}{2}$ and $R = 10 \text{ cm}$,the image is obtained at a distance from $P$ equal to:

Three glass cylinders of equal height $H = 30 \text{ cm}$ and same refractive index $n = 1.5$ are placed on a horizontal surface as shown in the figure. Cylinder $I$ has a flat top,cylinder $II$ has a convex top,and cylinder $III$ has a concave top. The radii of curvature of the two curved tops are same $(R = 3 \text{ m})$. If $H_1, H_2$ and $H_3$ are the apparent depths of a point $X$ on the bottom of the three cylinders,respectively,the correct statement$(s)$ is/are:
$(1) H_3 > H_1$
$(2) 0.8 \text{ cm} < (H_2 - H_1) < 0.9 \text{ cm}$
$(3) H_2 > H_3$
$(4) H_2 > H_1$

The figure shows a transparent sphere of radius $R$ and refractive index $\mu$. An object $O$ is placed at a distance $x$ from the pole of the first surface so that a real image is formed at the pole of the exactly opposite surface. If the refractive index $\mu$ of the sphere is varied,then the position $x$ of the object will also vary. Identify the correct statement.

$A$ spherical surface of radius of curvature $R$ separates air from glass of refractive index $1.5$. The centre of curvature is in the glass. $A$ point object $P$ placed in air forms a real image $Q$ in the glass. The line $PQ$ cuts the surface at point $O$ and $PO = OQ = x$. Hence the distance $x$ is equal to (in $R$)

An image is formed at a distance of $100 \ cm$ from the glass surface when light from a point source in air falls on a spherical glass surface with a refractive index of $1.5$. The distance of the light source from the glass surface is $100 \ cm$. The radius of curvature is (in $cm$)

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