Optics – NEET UG PhysicsPractice Questions & PYQs

When unpolarized light strikes a transparent medium at the polarizing angle such that the reflected and refracted rays are mutually perpendicular, the reflected light becomes completely polarized. The transverse nature of light waves ensures that the oscillating electric field vector in this reflected beam has no component parallel to the plane of incidence, leaving only vibrations perpendicular to it.

The angular fringe width in a double-slit setup is determined by the ratio of the wavelength to the slit separation, expressed as $\theta = \frac{\lambda}{d}$. Since the wavelength and screen distance remain constant, the angular width is inversely proportional to the slit separation, meaning their product remains unchanged. Equating the product of the initial angular width and initial separation to the product of the new angular width and unknown separation yields the required new distance of 1.9 mm.

The positions of the object and image relative to a curved mirror are connected through the mirror equation $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$. Substituting the initial values of focal length -15 cm and object distance -40 cm yields an initial image distance of -24 cm. Shifting the object 20 cm closer updates the object distance to -20 cm, resulting in a new image distance of -60 cm. The final image displacement is the difference between the two image locations, showing a movement of 36 cm away from the mirror.

When unpolarized light passes through the first polarizer, its intensity is reduced by half according to the polarization property, yielding an initial transmitted intensity of $\frac{I_0}{2}$. As this polarized light encounters the intermediate polarizer placed at an angle of $45^\circ$, Malus's Law dictates that the intensity scales by $\cos^2(45^\circ)$, dropping the value to $\frac{I_0}{4}$. Finally, since the third polarizer is perpendicular to the first, its angle relative to the intermediate polarizer is also $45^\circ$, applying the cosine-squared reduction once more to arrive at the final transmitted intensity of $\frac{I_0}{8}$.

Achieving optical dispersion without deviation requires the total deviation produced by the individual prisms to cancel each other out completely. Equating the deviation of the first prism to that of the second prism establishes the relation $A_1(\mu_1 - 1) = A_2(\mu_2 - 1)$. Substituting the given values into this expression creates a straightforward linear equation for the unknown refracting angle. Solving for the second prism angle results in $6^\circ$.

The resolving power of an optical microscope is inversely proportional to the wavelength of the illuminating light, meaning shorter wavelengths yield finer detail distinction. Expressing this relationship mathematically, the ratio between the resolving powers for two different wavelengths becomes $\frac{RP_1}{RP_2} = \frac{\lambda_2}{\lambda_1}$. Substituting the given wavelength values into this ratio produces $\frac{6000}{4000}$, which readily simplifies to $3/2$.

When Young's double slit experiment is immersed in a refractive medium, the wavelength of light decreases by a factor equal to the refractive index, which shifts the fringe positions accordingly. Equating the spatial position of the eighth bright fringe in the medium to the position of the fifth dark fringe in air sets up the fundamental equality $\frac{8 \lambda_m D}{d} = \frac{(2(5) - 1) \lambda D}{2 d}$. Substituting the medium wavelength relation $\lambda_m = \frac{\lambda}{\mu}$ into this expression allows the wavelength and apparatus dimensions to cancel out. Rearranging the resulting algebraic equation yields $\mu = \frac{16}{9}$, which evaluates to approximately $1.78$.

Rotating a plane mirror by an angle causes the reflected light ray to deflect by twice that rotation angle due to the geometry of reflection. For a source and scale situated at a distance $x$ from the mirror, the linear displacement $y$ of the light spot on the scale relates to the total deflection angle through the arc length relation. Equating the linear displacement to the product of the distance and the doubled angle gives $2\theta x = y$. Isolating the rotation angle from this expression directly results in $\frac{y}{2x}$.

Maximum and minimum intensities in an interference pattern depend on the square root of the individual source intensities, expressed through the relations $I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2$ and $I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2$. Substituting the given intensity ratio where $I_1 = n I_2$ allows factoring out the common intensity term to represent both extreme intensities entirely in terms of $n$. Forming the required ratio of the difference to the sum of these intensities and simplifying the algebraic expression yields $\frac{2\sqrt{n}}{n+1}$.

Single-slit diffraction produces dark bands on a screen at positions determined by the condition $y = \frac{n \lambda D}{a}$ for the $n$-th minima. For the first dark band, the integer $n$ equals one, allowing direct substitution of the given wavelength, screen distance, and aperture width. Inserting the values $\lambda = 5 \times 10^{-5}$ cm, $D = 60$ cm, and $a = 0.02$ cm into the formula gives $y = \frac{(5 \times 10^{-5}) \times 60}{0.02}$. Evaluating this expression yields a distance of $0.15$ cm from the center of the screen.

The separation between the objective and eyepiece lenses in a telescope is given by the tube length formula $L = v_o + f_e$, where $v_o$ is the image distance formed by the objective lens. Applying the thin lens equation $\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}$ to the objective lens with an object distance of $-200 \text{ cm}$ and a focal length of $40 \text{ cm}$ allows calculation of the image position. Substituting these values yields $\frac{1}{v_o} - \frac{1}{-200} = \frac{1}{40}$, which simplifies to $\frac{1}{v_o} = \frac{1}{40} - \frac{1}{200} = \frac{4}{200}$ and results in $v_o = 50 \text{ cm}$. Adding this image distance to the eyepiece focal length of $4 \text{ cm}$ gives a total lens separation of $54 \text{ cm}$.

The intensity distribution in Young's double slit experiment is expressed as $I = I_0 \cos^2\left(\frac{\pi y}{\beta}\right)$, where $\beta$ represents the fringe width. Given that the distance to the screen is $D = 10d$ and the slit separation is $d = 5\lambda$, the fringe width evaluates to $\beta = \frac{(10d)\lambda}{d} = 10\lambda$. A point directly in front of one of the slits is located at a transverse distance of $y = \frac{d}{2} = 2.5\lambda$ from the central axis. Substituting these values into the intensity formula gives $I = I_0 \cos^2\left(\frac{\pi(2.5\lambda)}{10\lambda}\right)$, which simplifies to $I_0 \cos^2\left(\frac{\pi}{4}\right)$ and yields an intensity of $\frac{I_0}{2}$.

When a light ray passes through a prism under minimum deviation conditions, the angle of incidence equals the angle of emergence, and the refractive index is determined by the relation $\mu = \frac{\sin\left(\frac{\delta_m + A}{2}\right)}{\sin\left(\frac{A}{2}\right)}$. Since the angle of incidence is given as $45^{\circ}$ and minimum deviation implies symmetry where $i = e$, the angle of minimum deviation is calculated using $\delta_{\min} = i + e - A$, which evaluates to $45^{\circ} + 45^{\circ} - 60^{\circ} = 30^{\circ}$. Substituting the angle of minimum deviation and the prism angle $A = 60^{\circ}$ into the refractive index formula yields $\frac{\sin(45^{\circ})}{\sin(30^{\circ})}$, simplifying to $\frac{1/\sqrt{2}}{1/2} = \sqrt{2}$.

The apparent depth of an object inside a denser medium is equal to its real depth divided by the refractive index. When viewed from opposite faces, the sum of the two apparent depths equals the total thickness of the slab divided by the refractive index. This relationship is expressed as $\frac{l}{\mu} = 3 + 5$, which simplifies to $\frac{l}{\mu} = 8$. Multiplying this combined result by the refractive index of $1.5$ gives a total slab thickness of $12$ cm.

Diffraction minima occur when the path difference satisfies $a \sin\theta = n\lambda$, while secondary maxima appear where the path difference is approximately $a \sin\theta' = \left(n + \frac{1}{2}\right)\lambda$. For the first minimum at $30^{\circ}$, substituting $n = 1$ gives $a \sin(30^{\circ}) = \lambda$, which reveals that the slit width equals two wavelengths, or $a = 2\lambda$. Applying this condition to the first secondary maximum where $n = 1$, the relation becomes $(2\lambda) \sin\theta' = \frac{3\lambda}{2}$. Solving this equation yields $\sin\theta' = \frac{3}{4}$, resulting in an angle of $\sin^{-1}\left(\frac{3}{4}\right)$.

The equivalent focal length of a combination of lenses in contact is found using the inverse summation formula $\frac{1}{f_{\text{eq}}} = \sum \frac{1}{f_i}$, where individual focal lengths are derived from the lens maker's equation. For the equiconvex glass lenses with refractive index $\frac{3}{2}$, the focal length satisfies $\frac{1}{f} = \left(\frac{3}{2} - 1\right)\left(\frac{2}{R}\right) = \frac{1}{R}$, implying that $R = f$. The water film trapped between the lenses acts as a biconcave lens with refractive index $\frac{4}{3}$, having a focal length $\frac{1}{f'} = \left(\frac{4}{3} - 1\right)\left(-\frac{2}{R}\right) = -\frac{2}{3f}$. Combining the two glass lenses and the central water lens gives $\frac{1}{f_{\text{eq}}} = \frac{1}{f} - \frac{2}{3f} + \frac{1}{f} = \frac{4}{3f}$, which results in an equivalent focal length of $\frac{3f}{4}$.

The sign and magnitude of magnification produced by spherical mirrors dictate the orientation, nature, and size of the resulting image, where negative values correspond to real and inverted images and positive values indicate virtual and erect ones. A magnification of $m = -2$ represents an inverted, enlarged, and real image, characteristics exclusively formed by a concave mirror. Similarly, a magnification of $m = -\frac{1}{2}$ describes an inverted, diminished, and real image, which is also produced by a concave mirror. In contrast, a magnification of $m = +2$ yields an erect, enlarged, and virtual image, uniquely forming with a concave mirror. Finally, a magnification of $m = +\frac{1}{2}$ corresponds to an erect, diminished, and virtual image, which is the defining behavior of a convex mirror.

The thin lens formula relates object distance, image distance, and focal length to correct vision abnormalities. To shift the far point from a finite distance to infinity, the corrective lens must form a virtual image of a distant object at the person's actual maximum viewing distance of $-4$ m. Substituting these values into the lens equation $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ yields a focal length of $-4$ m. The optical power of a lens is the reciprocal of its focal length in meters, resulting in a power of $-0.25$ diopters achieved using a concave lens.

The angular width of the central maximum in a single-slit diffraction pattern is given by $\frac{2\lambda}{d'}$, where $d'$ is the width of the slit. Meanwhile, the angular span occupied by ten interference maxima in a double-slit arrangement is determined by $10 \left(\frac{\lambda}{d}\right)$, where $d$ is the separation between the slits. Equating these two angular extents because the ten double-slit maxima fit exactly within the central single-slit maximum gives $\frac{2\lambda}{d'} = \frac{10\lambda}{d}$. Solving this relation yields $d' = 0.2 d$, and substituting the given slit separation of 1 mm results in a required slit width of 0.2 mm.

The magnifying power of an astronomical telescope in normal adjustment is defined as the ratio of the focal length of the objective to that of the eyepiece, expressed as $M = \frac{f_0}{f_e}$. When a black line of length $L$ is marked on the objective lens, it acts as an object placed at a distance equal to the tube length $f_0 + f_e$ from the eyepiece. The linear magnification produced by the eyepiece relates the size of the real image $I$ to the object size $L$ through the lens configuration, yielding the ratio $\frac{I}{L} = \frac{f_e}{f_0}$. Inverting this ratio directly connects the image dimensions to the optical magnification, proving that the magnifying power equals $\frac{L}{I}$.