Formation of real image using a biconvex lens is shown below: JEE Main 2019 Physics, Ray Optics — question figure If the whole set up is immersed in…

For a transparent prism, if the angle of minimum deviation is equal to its refracting angle, the refractive index $n$ of the prism satisfies.

A parallel beam of light travelling in air (refractive index 1.0) is incident on a convex spherical glass surface of radius of curvature 50 cm. Refractive index of glass is 1.5. The rays converge to a point at a distance $x \mathrm{~cm}$ from the centre of the curvature of the spherical surface. The value of $x$ is $\_\_\_\_$ cm.

A thin convex lens of focal length 5 cm and a thin concave lens of focal length 4 cm are combined together (without any gap) and this combination has magnification $m_{1}$ when an object is placed 10 cm before the convex lens. Keeping the positions of convex lens and object undisturbed a gap of 1 cm is introduced between the lenses by moving the concave lens away, which lead to a change in magnification of total lens system to $m_{2}$. The value of $\left|\frac{m_{1}}{m_{2}}\right|$ is $\_\_\_\_$.

As shown in the diagram, when the incident ray is parallel to base of the prism, the emergent ray grazes along the second surface.  If refractive index of the material of prism is $\sqrt{2}$, the angle $\theta$ of prism is.

The size of the images of an object, formed by a thin lens are equal when the object is placed at two different positions 8 cm and 24 cm from the lens. The focal length of the lens is $\_\_\_\_$ cm.