NEET Ray Optics MCQs & PYQs

For total internal reflection, the incidence angle must be greater than the critical angle, so $\sin i > \sin \theta_c = \frac{1}{\mu}$. Given $i = 45^\circ$, we have $\frac{1}{\sqrt{2}} > \frac{1}{\mu} \implies \mu > \sqrt{2} \approx 1.414$. Among the options, only $\mu = 1.50$ satisfies this.

The light is cut off if the disc covers the area defined by the critical angle. Radius is $r = \frac{h}{\sqrt{\mu^2 - 1}}$. Substituting $h = 4 m$ and $\mu = \frac{5}{3}$, $r = \frac{4}{\sqrt{(\frac{5}{3})^2 - 1}} = \frac{4}{\frac{4}{3}} = 3 m$.

Optical fibers work on the principle of total internal reflection. Light entering the fiber strikes the core-cladding interface at an angle greater than the critical angle, reflecting entirely within the core.

For light to undergo total internal reflection at the lateral surface regardless of the incidence angle at the entrance face, the critical angle must be less than or equal to $45^\circ$. Thus, $\sin \theta_c = \frac{1}{n} \le \sin(45^\circ) = \frac{1}{\sqrt{2}}$, which gives $n > \sqrt{2}$.

Radius of the disc is $r = \frac{h}{\sqrt{\mu^2 - 1}} = \frac{4}{\sqrt{(\frac{5}{3})^2 - 1}} = \frac{4}{4/3} = 3 m$. The minimum diameter of the disc required is $d = 2r = 2(3) = 6 m$.

A spherometer is an instrument specifically designed to measure the radius of curvature of spherical surfaces. Therefore, it can be used to measure the radius of curvature of the curved surface of the plano-convex lens.

The focal length of a lens in a liquid is given by $1/f_l = (\frac{\mu_g}{\mu_l} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$. When the refractive indices are equal ( $\mu_l = \mu_g$ ), the term $(\frac{\mu_g}{\mu_l} - 1)$ becomes zero. This makes $1/f_l = 0$, meaning the focal length becomes infinite.

According to Cauchy's relation, the refractive index is inversely proportional to wavelength, so $\mu_v > \mu_r$.[cite: 1] From the Lens Maker's formula, the focal length is inversely proportional to $(\mu - 1)$. Since violet has a higher refractive index than red, its focal length will be shorter, hence $f_v < f_r$.[cite: 1]

Using the relation $f_l / f_a = (\mu_g - 1) / (\mu_g / \mu_l - 1)$, we substitute the given values: $f_l / 2 = (1.5 - 1) / (1.5 / 1.25 - 1)$.This simplifies to $f_l / 2 = 0.5 / (1.2 - 1) = 0.5 / 0.2 = 2.5$. Therefore, the focal length in liquid is $f_l = 2.5 \times 2 = 5 cm$.

For the emergent beam to be parallel, the image formed by the convex lens must lie exactly at the focal point of the concave lens. The convex lens focuses the parallel beam at $+20 cm$. The concave lens requires a virtual object at its focus ($+5 cm$ behind it) to collimate the rays. Thus, $d + 5 = 20$, which gives $d = 15 cm$.