Optics – NEET UG PhysicsPractice Questions & PYQs

According to Snell's law, the angle of refraction depends on the refractive index of the medium, which varies slightly with wavelength for different colors. Since red and green rays travel with different speeds inside the glass, they refract at slightly different angles and thus separate spatially inside the slab. However, because the entry and exit faces of a rectangular glass slab are parallel, the net deviation cancels out upon emergence, causing both rays to emerge from two distinct points while propagating in parallel directions.

The focal length of a lens is determined by the Lens maker's formula, expressed as $\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$, which connects the refractive index and the radii of curvature. When the equiconvex lens is sliced horizontally along $XOX'$, each resulting piece retains its original curved boundaries, meaning the focal length remains unchanged so that $f' = f$. Conversely, slicing the lens vertically along $YOY'$ creates plano-convex pieces where one surface becomes entirely flat, making its radius of curvature equal to infinity. Substituting this flat surface into the formula gives $\frac{1}{f''} = (\mu - 1)\left(\frac{1}{R} - \frac{1}{\infty }\right)$, which doubles the focal length to $f'' = 2f$.

The behavior of a lens in any surrounding medium is governed by the lens maker formula $\frac{1}{f} = \left(\frac{n_{\text{lens}}}{n_{\text{liquid}}} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$. When a convex lens is immersed in a liquid whose refractive index matches that of the lens material, the relative refractive index term becomes unity. Consequently, the optical power vanishes, causing light rays to pass straight through without bending, which corresponds to an infinite focal length.

A convex lens forms a real image of an object on a screen when the total minimum separation between the object and the screen is equal to four times the focal length, placing the lens at the exact midpoint where object and image distances are equal. Since the lens is positioned at a distance $d$ from the second wall, this distance represents both the image distance and half of the total separation between the two walls. Therefore, the required focal length is equal to half of this distance, which evaluates to $d/2$.

Total internal reflection requires the angle of incidence inside the medium to meet or exceed the critical angle, relying directly on the principles of refraction and boundary geometry. At the point of entry on the first face of the prism, applying Snell's law with an incidence angle of $45^{\circ}$ determines the angle of refraction such that $\sin r = \frac{1}{\sqrt{2} \mu}$. Upon reaching the second boundary, the ray meets the condition for total internal reflection, where the critical angle is defined by $\sin i_c = \frac{1}{\mu}$. The internal geometry dictates that the sum of this refraction angle and the critical angle equals $90^{\circ}$, making $\sin r = \cos i_c$. Equating these expressions leads to $\frac{1}{\sqrt{2} \mu} = \frac{\sqrt{\mu^2 - 1}}{\mu}$, which simplifies to give the required refractive index of $\mu = \sqrt{\frac{3}{2}}$.

The minimum resolvable linear distance between two points is determined by multiplying the distance from the observer by the angular resolution limit of the eye lens. This angular limit is given by the ratio of the wavelength of light to the diameter of the aperture. Converting all dimensions to centimeters, the separation is calculated by multiplying the distance of 5000 cm by the wavelength of 5000 x 10^-8 cm, and then dividing by the lens diameter of 2 x 10^-1 cm. This computation evaluates to a minimum resolvable distance of 1.25 cm.

The frequency of a light wave is determined solely by the source and remains invariant when the wave crosses the boundary between two different optical media. Although the velocity, wavelength, and intensity change as light enters the water, the frequency $n = n'$ does not change.

The boundary of the illuminated circle on the water surface is defined by the critical angle of incidence, where light rays are refracted at an angle of 90 degrees and cannot escape into the air. Using Snell's law, the critical angle satisfies the relation $\sin \theta = \frac{1}{\mu}$, which evaluates to $3/5$ given a refractive index of $5/3$. From the geometry of the light cone, the radius of the disc relates to depth through the expression $r = h \tan \theta$. Substituting a depth of 4 meters and a tangent value of $3/4$, the minimum radius becomes 3 meters.

Optical fibres rely on the principle of total internal reflection, which occurs when a light ray traveling within a denser core strikes the boundary of a surrounding cladding with a lower refractive index at an angle exceeding the critical angle. Because the angle of incidence is consistently greater than this threshold, the light undergoes repeated reflection rather than refraction, trapping the beam inside the core. This mechanism allows optical signals to propagate over long distances with minimal energy loss or attenuation.

The formation of a rainbow relies on the dual action of refraction, dispersion, and internal reflection as sunlight passes through spherical water droplets suspended in the atmosphere. When white light enters a droplet, it refracts and splits into its component wavelengths due to varying degrees of bending, a process known as dispersion. These dispersed rays then reflect off the back inner surface of the droplet before refracting out into the air toward an observer.

A silvered lens functions optically as a combination of a lens refracting light twice and a mirror reflecting it from the coated surface. The focal length of the unsilvered plano-convex lens is first determined using the lens maker's formula $\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$, which results in a lens focal length of 20 centimeters for a radius of 10 centimeters and a refractive index of 1.5. Since the flat surface is silvered, it acts as a plane mirror with infinite focal length, making the total optical power equivalent to twice the lens power. Consequently, the final focal length after silvering is given by $F = \frac{f}{2}$, yielding a value of 10 centimeters.

Total internal reflection ensures that light striking the lateral boundary at an angle greater than the critical angle remains trapped inside the medium regardless of the initial entry angle. When light enters from air at the maximum possible angle of incidence of $90^\circ$, the angle of refraction inside the rod reaches the critical angle $C$, meaning $\sin C = \frac{1}{n}$. The geometric relation at the internal boundary requires the angle of incidence against the lateral wall to be $90^\circ - C$, so total internal reflection occurs if $\sin(90^\circ - C) \ge \frac{1}{n}$, which simplifies to $\cos C \ge \frac{1}{n}$. Substituting the trigonometric identity $\cos C = \sqrt{1 - \frac{1}{n^2}}$ into the inequality yields $\sqrt{1 - \frac{1}{n^2}} \ge \frac{1}{n}$, which ultimately leads to $n > \sqrt{2}$ for strict containment.

The refractive index of a prism is related to its angle of minimum deviation through the prism formula $\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$. Substituting the equilateral prism angle of 60 degrees and the refractive index value of $\sqrt{3}$ into the equation gives $\sqrt{3} = \frac{\sin\left(\frac{60^{\circ} + \delta_m}{2}\right)}{\sin(30^{\circ})}$, which simplifies to $\frac{\sqrt{3}}{2} = \sin\left(\frac{60^{\circ} + \delta_m}{2}\right)$. Solving this trigonometric relation results in an angle of minimum deviation of 60 degrees.

A light ray retraces its path through a polished prism when it strikes the silvered surface normally, making the internal angle of refraction equal to the prism angle. Applying Snell's law, expressed as $\mu = \frac{\sin i}{\sin r}$, the refractive index is multiplied by the sine of this internal angle. This calculation results in a sine value of 0.707, corresponding to an angle of incidence of 45 degrees.

When an object inside a transparent medium is viewed from opposite sides, the sum of the two apparent depths equals the total apparent thickness perceived by the observers. The true thickness of the slab acts as the real depth, which is related to the total apparent depth $y$ by the refractive index through the formula $\mu = \frac{x}{y}$. Adding the two apparent distances of 5 centimeters and 2 centimeters gives a total apparent depth of 7 centimeters. Multiplying this sum by the refractive index of 1.5 yields a total slab thickness of 10.5 centimeters.

The displacement method relies on the conjugate foci property of lenses, establishing that the actual distance between virtual sources equals the geometric mean of the two measured separations. Applying the formula $d = \sqrt{d_1 d_2}$ with the given values yields $d = \sqrt{16 \times 9}$, which evaluates to an actual separation of 12 cm.

According to the displacement method principles for a lens, the linear size of an object is equal to the geometric mean of the linear sizes of its two conjugate images, expressed as $O = \sqrt{I_{1}I_{2}}$. Since the area of a shape scales with the square of its linear dimensions, the area of the light source can be derived by applying the geometric mean property directly to the areas. Consequently, the area of the source is given by the square root of the product of the two image areas, represented as $\sqrt{A_{1}A_{2}}$.

A light beam entering a prism with a silvered back surface will retrace its path upon reflection if it strikes the mirrored surface normally, which implies that the angle of refraction inside the prism equals the refracting angle. Applying Snell's law at the entry face gives the relation $\sin i = \mu \sin r$, where $r$ matches the prism angle of 30°. Substituting the refractive index of $\sqrt{2}$ and $\sin 30^\circ$ into $\sin i = \sqrt{2} \times \sin 30^\circ$ results in $\sin i = \frac{1}{\sqrt{2}}$, yielding an angle of incidence of 45°.

The geometry of reflection in a plane mirror dictates that the angle of incidence equals the angle of reflection, causing the vertical span of the mirror to intercept light symmetrically. Through similar triangles formed by the light paths from the top and bottom of a person to the eyes, the required viewing height is precisely halved. Therefore, for a person standing 6 feet tall, the minimum length of the mirror needed to capture the entire reflection is 3 feet.