Optics – NEET UG PhysicsPractice Questions & PYQs

The absolute refractive index of a medium is inversely proportional to the speed of light within it, expressed as $\mu = \frac{c}{v}$, and the critical angle for total internal reflection satisfies Snell's law as $\mu_A \sin \theta_c = \mu_B \sin 90^\circ$. Dividing the speed of light in vacuum by the given speeds in media A and B yields refractive indices of 2 and 1.5, respectively. Substituting these refractive indices into the boundary equation isolates the sine of the critical angle as 0.750, making the critical angle equal to $\sin^{-1}(0.750)$.

The fringe width in a Young's double-slit setup is expressed by the relation $\beta = \frac{\lambda D}{d}$, which shows that the linear separation between consecutive fringes is directly proportional to the distance of the screen from the slits. When the screen is moved further away, the value of the distance parameter increases, causing a proportional enlargement of the overall fringe pattern. Consequently, the linear separation of the fringes expands across the screen.

When light strikes a transparent boundary at an angle of incidence whose tangent equals the refractive index, a condition known as Brewster's law, the reflected and refracted rays become mutually perpendicular. Calculating the inverse tangent of the given refractive index of $\sqrt{3}$ yields an angle of $60^\circ$, which precisely matches the provided angle of incidence. Consequently, the angle formed between the reflected and refracted components is $90^\circ$.

The behavior of light as it crosses the boundary of the prism is governed by Snell's law, which relates the angle of incidence to the angle of emergence using the refractive indices of the involved mediums. Applying this principle at the boundary gives the equation $\frac{\sin 30^\circ}{\sin e} = \frac{1}{\sqrt{3}}$ where $e$ is the angle of emergence and the glass has a refractive index of $\sqrt{3}$.

Isolating the sine of the unknown angle yields $\sin e = \sqrt{3} \times \sin 30^\circ$. Substituting the standard trigonometric value transforms this into $\sin e = \sqrt{3} \times \frac{1}{2}$, which simplifies to $\frac{\sqrt{3}}{2}$. Evaluating this ratio reveals that the angle of emergence is $60^\circ$.

The optical performance of an astronomical telescope objective relies heavily on its physical dimensions, where a larger aperture directly increases light-gathering power to capture dim celestial objects. Furthermore, resolving power improves with a wider diameter, allowing astronomers to clearly separate closely spaced stars or planetary features. Consequently, these combined enhancements in resolution, brightness, and overall image visibility make a large aperture indispensable for high-quality astronomical observations.

When a parallel beam of light strikes a convex lens, it converges toward the focal point of that lens, but the presence of a subsequent concave lens intercepts this light and bends it back into a parallel path. For the emerging rays to become parallel, the converging rays from the convex lens must be directed precisely toward the virtual focus of the concave lens. Therefore, the distance between the two lenses is determined by subtracting the focal length of the concave lens from the focal length of the convex lens, yielding $20\text{ cm} - 5\text{ cm} = 15\text{ cm}$.

The lens formula, represented as $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$, governs the refraction of light through spherical lenses. For the initial refraction through the convex lens, substituting the object distance of 60 cm and focal length of 30 cm places the first image at a distance of 60 cm beyond the lens.

Because the plane mirror stands 40 cm away from the lens, this image lies 20 cm in front of the reflecting surface, which in turn creates a reflected image positioned 20 cm behind the mirror. This reflected position serves as a new source for a second refraction through the lens, corresponding to an object distance of 20 cm. Applying the lens formula for this return path yields a final image distance of 60 cm on the side opposite the lens, placing it 20 cm away from the plane mirror and rendering it a virtual image.

When a light ray emerges normally from the second surface of a prism, the angle of emergence and the second refracting angle are both zero. Using the geometric relation for a prism where the sum of the internal refracting angles equals the prism angle, the first refracting angle simplifies to $A$. Applying Snell's law at the initial boundary gives $\sin i = \mu \sin A$, and for a small angle prism, approximating the sine of an angle as the angle itself yields $i = \mu A$.

The fringe width in a double-slit setup is determined by the formula $\beta = \frac{\lambda D}{d}$, where $\lambda$ is the wavelength, $D$ is the screen distance, and $d$ is the source separation. Substituting the modified parameters into this expression yields $\beta' = \frac{\lambda (2D)}{d / 2}$. Simplifying this rearranged fraction results in $\beta' = \frac{4\lambda D}{d}$, which shows that the new fringe width becomes four times the original value.

Brewster's law connects the refractive index of an optical medium to the polarizing angle through the relation $\mu = \tan i_b$. Since any practical transparent medium has a refractive index strictly greater than unity, the tangent of the Brewster angle must likewise exceed one. Applying the inverse tangent function across this range establishes that the polarizing angle falls strictly within the interval $45^{\circ} < i_b < 90^{\circ}$.

The angular resolution limit of an optical telescope is determined by the diffraction formula $\theta = \frac{1.22 \lambda}{d}$, which relates the operating wavelength to the diameter of the objective lens. Substituting the given values for the wavelength and the aperture diameter into the expression gives $\frac{1.22 \times 600 \times 10^{-9}\text{ m}}{2}$. Simplifying this calculation results in an angular limit of resolution equal to $3.66 \times 10^{-7}\text{ rad}$.

Within the visible light spectrum, colours are arranged in order of increasing frequency and decreasing wavelength from red to violet. Red light resides at the extreme lower end of the visible frequency band, possessing the minimum amount of energy per photon among these options. Consequently, this lower frequency directly corresponds to the longest physical wavelength in the visible spectrum.

The combined focal length of thin lenses placed in contact is derived by summing their individual optical powers according to the lens maker's principle. For two identical equi-convex lenses in contact in air, the reciprocal of the equivalent focal length $F_1$ is given by $\frac{1}{F_{1}} = \frac{1}{f} + \frac{1}{f} = \frac{2}{f}$. When the intervening space is filled with a liquid matching the refractive index of the glass, it introduces a diverging liquid lens with a focal length of $-f$, modifying the net inverse focal length to $\frac{1}{F_{2}} = \frac{1}{f} + \frac{1}{f} - \frac{1}{f} = \frac{1}{f}$. Comparing these reciprocal values yields a focal length ratio of 1 to 2.

The angular width of interference fringes scales inversely with the refractive index of the surrounding medium because light experiences a proportional reduction in wavelength upon immersion. The relationship governing this shift is given by $\theta_{w} = \frac{\theta_{0}}{\mu}$, where the angular width in air is divided by the refractive index of the new medium. Evaluating this formula with the initial air measurement and the refractive index of water results in an angular width of 0.15 degrees.

Snell's law governs the refraction of light as it passes between different optical media, establishing a specific threshold known as the critical angle where the angle of refraction reaches its maximum possible boundary value. When a light ray traveling within a denser medium strikes the interface at this precise critical angle, it fails to emerge into the rarer medium and instead travels directly along the boundary surface. Consequently, the angle formed between the normal and this surface-grazing refracted ray measures exactly 90 degrees.

Destructive interference at the dark fringes of a double-slit setup corresponds to a path difference given by the general expression $\Delta x = (2n - 1)\frac{\lambda}{2}$. Substituting the fifth order into this relation where $n = 5$ gives $(2(5) - 1)\frac{\lambda}{2}$, which immediately evaluates to $\frac{9\lambda}{2}$.

The optical power of a lens is determined by its focal length, which depends entirely on the radii of curvature of its surfaces and the relative refractive index of the lens material. Cutting an equiconvex lens longitudinally along a plane containing the principal axis divides its aperture area in half, but leaves the curvature of both the entry and exit surfaces completely unchanged. Because the focal length of each resulting piece remains identical to the original lens, the optical power of the individual half also remains unchanged at $P$.

Atmospheric optics dictates that a rainbow is formed by the combined dispersion, refraction, and internal reflection of sunlight passing through water droplets suspended in the air. Because these optical phenomena require sunlight to enter the back of the droplets and reflect toward the observer's eyes at precise angular intervals, the sun must always be positioned directly behind the person viewing the display. Consequently, an observer can never witness a rainbow while facing toward the sun, making that specific statement incorrect.

The performance of an astronomical refracting telescope depends on maximizing both its magnifying power and its ability to distinguish fine details, which are governed by separate optical parameters. Because angular magnification is determined by the ratio $\frac{f_0}{f_e}$, a large objective focal length $f_0$ is necessary to scale up distant images. Concurrently, angular resolution is dictated by the diffraction limit expression $\frac{D}{1.22\lambda}$, meaning a large objective diameter $D$ is required to resolve closely spaced celestial objects. Therefore, achieving both optimal magnification and high clarity mandates an objective lens characterized by an extensive focal length and a wide aperture.

A light ray retraces its path after hitting a silvered surface inside a prism only when it strikes that surface at normal incidence, meaning the angle of refraction inside the prism must equal the prism angle of 30°. Applying Snell's law at the entry face with a refractive index of $\sqrt{2}$ sets up the relation $\frac{\sin i}{\sin 30^\circ} = \sqrt{2}$. Solving this gives $\sin i = \sqrt{2} \times \frac{1}{2} = \frac{1}{\sqrt{2}}$, which corresponds to an angle of incidence of 45°.