Optics – NEET UG PhysicsPractice Questions & PYQs

The behavior of light inside a prism is determined by Snell's law and the geometric constraints of the refracting surfaces. At the point of entry $P$, applying Snell's law relates the incoming angle to the interior refraction angle through the equation $1 \cdot \sin(30^{\circ}) = \mu \sin(r_1)$.

Since the ray travels parallel to the base and emerges cleanly along the face $AC$, the angle of incidence at the second boundary matches the critical angle $c$. For a right-angled prism, geometry dictates that $r_1 = 90^{\circ} - c$, which transforms Snell's equation using the trigonometric identity $\sin(90^{\circ} - c) = \cos(c) = \frac{\sqrt{\mu^2 - 1}}{\mu}$. Substituting this back gives $\frac{1}{2} = \sqrt{\mu^2 - 1}$, and squaring both sides results in $\frac{1}{4} = \mu^2 - 1$, simplifying directly to $\mu = \frac{\sqrt{5}}{2}$.

When unpolarized light strikes a transparent dielectric surface at Brewster's angle, the reflected and refracted rays become mutually perpendicular. At this specific angle of incidence, the vibrations parallel to the plane of incidence are completely refracted, leaving the reflected ray entirely polarized with vibrations perpendicular to the incidence plane. Conversely, the transmitted refracted ray contains a mix of both polarization states, making it only partially polarized.

The magnifying power of an astronomical telescope for viewing a distant object is defined as the ratio of the focal length of the objective lens to the focal length of the eyepiece. Substituting the given focal length values into the formula yields $m = \frac{f_0}{f_e}$. Evaluating this gives $m = \frac{140}{5}$, which simplifies to $28$.

Image formation relies on the coherent preservation of light rays through regular reflection or refraction, which allows light from an object to converge at a specific point, distinguishing mirrors from rough walls. Furthermore, the diverse colors perceived in our surroundings arise because white light is composed of a spectrum of constituent wavelengths that interact differently with various objects through selective absorption and reflection. Therefore, both statements regarding image optics and light composition are entirely accurate.

When white light replaces a monochromatic source in Young's double slit experiment, the path difference at the central point on the screen is zero for all component wavelengths simultaneously. This causes all constituent colors to constructively interfere at the center, producing a bright white fringe. Since fringe width depends directly on wavelength according to the relation $\beta = \frac{\lambda D}{d}$, the surrounding maxima for different colors spread out at slightly different positions, creating a few distinct colored fringes before overlapping.

When unpolarized light passes through the first polaroid, its intensity is immediately halved, yielding an initial transmitted intensity of $I_1 = \frac{I_0}{2}$. As this polarized beam subsequently encounters polaroid B oriented at an angle of $60^\circ$, the intensity changes according to Malus's law, resulting in an intensity of $I_B = \frac{I_0}{2}\cos^2(60^\circ)$, which evaluates to $\frac{I_0}{8}$. Finally, passing this light through polaroid C set at $45^\circ$ relative to B yields the ultimate emergent intensity of $I_C = \frac{I_0}{8}\cos^2(45^\circ)$, successfully reducing the final value to $\frac{I_0}{16}$.

The angular fringe width in a double-slit interference pattern depends exclusively on the wavelength of light and the slit separation, remaining completely independent of the distance between the screen and the slits. Consequently, moving the screen away alters neither of these core parameters, keeping the angular separation constant and verifying the first statement. Meanwhile, increasing the wavelength of the monochromatic light source directly increases the angular separation rather than decreasing it, which invalidates the second statement.

A convex lens produces a real image of the exact same size as the object only when both the object distance and the image distance equal twice the focal length. Since the lens is positioned at a distance $x$ from the second wall where the image is formed, the image distance equals $x$, meaning $x = 2f$. Rearranging this relationship yields a required focal length of half that distance, or $\frac{x}{2}$.

The refraction of light through any lens depends directly on the refractive index of the material it is made from. Since the lens is constructed from three different transparent media, it possesses three distinct refractive index values along its structure. When light rays from a point object placed beyond $2f$ encounter these sections, each portion bends the light independently according to its specific optical density. Consequently, the device behaves like three separate lenses operating simultaneously, causing the rays to converge at three distinct locations to form three separate real images on the opposite side.

The refractive index of a medium is defined as the ratio of the speed of light in a vacuum to its speed in that medium, which can be expressed in terms of electric permittivity and magnetic permeability. Substituting the given values of $2 E_0$ for permittivity and $1.5 \mu_0$ for permeability into the wave speed formula yields an expression where the free-space constants cancel out. Simplifying the resulting square root ratio directly produces a refractive index of $\sqrt{3}$.

The critical angle for a denser medium is determined by the inverse sine of the reciprocal of its refractive index, with the refractive index itself given by the ratio of the speed of light in air to the speed in the medium. The speed of light in air is represented by the distance $x$ divided by time $t_1$, whereas the speed in the denser medium is represented by the distance $10x$ divided by time $t_2$. Substituting these speeds into the refractive index ratio produces $\mu = \frac{t_2}{10 t_1}$, leading directly to a critical angle of $\sin^{-1} \left( \frac{10 t_1}{t_2} \right)$.

The critical angle for total internal reflection is governed by the relation $\sin i_{c} = \frac{1}{\mu}$, where the refractive index $\mu$ decreases as the wavelength $\lambda$ increases. Since the critical angle is directly proportional to the wavelength, colours with longer wavelengths exhibit larger critical angles. Therefore, the angle of incidence permits yellow, orange, and red light to cross the boundary and emerge into the air.

The behavior of light passing through a prism is governed by the geometric relation $A = r_1 + r_2$ and Snell's law expressed as $\sin e = \mu \sin r_2$. Since the incoming ray travels horizontally, it strikes the first face normally, making the initial angle of refraction zero ($r_1 = 0$). This causes the second angle of refraction to equal the entire prism angle of 6 degrees. Substituting these values into Snell's law with a refractive index of 1.5, and utilizing the small angle approximation where the sine of a small angle is nearly equal to the angle itself, the angle of emergence is found to be 9 degrees.

When multiple thin lenses are placed in close contact, the reciprocal of the equivalent focal length equals the algebraic sum of the reciprocals of the individual focal lengths. Combining a convex lens of positive focal length $f$ with a concave lens of equal magnitude negative focal length $-f$ results in a net sum of zero. Because the reciprocal of the equivalent focal length is zero, the equivalent focal length itself becomes infinite, meaning the combination behaves like a flat glass plate.

When thin lenses are placed in close contact, the system's combined optical power is determined by summing the reciprocal focal lengths of each component using the relation $\frac{1}{f_{\text{eff}}} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3}$. Each individual focal length is derived from the lens maker's formula $\frac{1}{f} = (\mu - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$. For the first boundary layer with a refractive index of 1.6, a flat incoming surface, and a 20 cm outgoing radius, the calculation gives $\frac{1}{f_1} = (1.6 - 1) \left(\frac{1}{∞} - \frac{1}{20}\right) = \frac{-0.6}{20}$. The central layer features a refractive index of 1.5 bounded by a 20 cm first radius and a -20 cm second radius, yielding $\frac{1}{f_2} = (1.5 - 1) \left(\frac{1}{20} - \frac{1}{-20}\right) = \frac{0.5}{10}$. The final layer mirrors the first with a refractive index of 1.6, a -20 cm first radius, and a flat exit surface, resulting \in $\frac{1}{f_3} = (1.6 - 1) \left(\frac{1}{-20} - \frac{1}{∞}\right) = \frac{-0.6}{20}$. Combining all three values produces $\frac{1}{f_{\text{eff}}} = \frac{-0.6}{20} + \frac{0.5}{10} - \frac{0.6}{20}$, which simplifies to $\frac{-0.1}{10}$ or $\frac{-1}{100}$, giving a final equivalent focal length of -100 cm.

The lens maker's formula, $\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$, connects a lens's focal length to its refractive index and curvature, while optical power is calculated as $P = \frac{1}{f}$. Substituting the refractive index of 1.5 and the radii of curvature of 20 cm into the formula gives a focal length of 20 cm, which converts to 0.2 meters. Taking the reciprocal of this focal length results in a lens power of +5 D.

According to Malus's law, the transmitted intensity of a linearly polarized beam passing through an analyzer depends on the square of the cosine of the angle between their respective transmission axes, given by the equation $I' = I \cos^2\theta$. Substituting the given angle into the formula yields $I' = I \cos^2(30^{\circ})$. Since the cosine of $30^{\circ}$ is $\frac{\sqrt{3}}{2}$, squaring this value gives $\frac{3}{4}$, which results in an emerging intensity of $\frac{3I}{4}$ after multiplying by the initial intensity.

The magnifying power of an astronomical telescope in normal adjustment is determined by the ratio of the objective focal length to the eyepiece focal length, and the final image formed is inverted rather than erect. The total tube length in this adjustment equals the sum of the focal lengths ($L = f_o + f_e$), yielding $20\text{ m} + 0.02\text{ m} = 20.02\text{ m}$, which confirms the first statement. The magnification is calculated by dividing the objective focal length by the eyepiece focal length with a negative sign, resulting in $-1000$. Since light passes through the entire optical system, the objective requires a larger light-gathering aperture than the eyepiece, making the fourth statement correct while rendering the third statement about an erect image false.

Rainbow formation is an optical phenomenon caused by the dispersion, refraction, and reflection of sunlight inside water droplets. A primary rainbow is produced by a single internal reflection and is observed at a lower angular elevation in the sky. In contrast, a secondary rainbow undergoes double internal reflection, which inverts its color order and positions it at a higher angular altitude directly above the primary arc.

The width of an interference fringe in a double-slit setup is expressed as $\beta = \frac{\lambda D}{d}$, and the total span across a fixed region of the screen remains constant regardless of the wavelength used. This means the product of the number of fringes and the fringe width stays invariant, leading to the relation $n_1 \lambda_1 = n_2 \lambda_2$. Substituting the initial 8 fringes with the 600 nm wavelength and the new 400 nm wavelength into this conservation equation determines the new fringe count to be 12.