Optics – NEET UG PhysicsPractice Questions & PYQs

In single-slit diffraction, the first minimum occurs when the total path difference between light waves originating from the top and bottom edges of the slit equals one full wavelength. Because the midpoint of the slit divides this total aperture in half, the path difference between a wavelet from the edge and a wavelet from the midpoint is exactly half of that, or $\frac{\lambda}{2}$. Converting this path difference into a phase difference using the standard relation $\phi = \frac{2\pi}{\lambda} \times \Delta x$ gives a value of $\pi$ radians.

Light rays successfully emerge from the prism only when the angle of incidence remains less than the critical angle, leading to the operational threshold $\mu \le \sqrt{2}$ derived from the condition $\sin i \le \sin C$. Comparing the refractive indices against this critical value of $1.414$ shows that the red wavelength has a refractive index of $1.39$, which is safely below the limit and allows it to transmit through the face. In contrast, both the green wavelength at $1.44$ and the blue wavelength at $1.47$ exceed this threshold, forcing them to undergo total internal reflection. This behavioral split results in the prism separating the red portion from the green and blue parts.

The angular spread of the central maximum in single-slit diffraction is defined by the angle subtended from the central axis to the first diffraction minimum, given by $\theta = \frac{\lambda}{a}$. The linear width on a screen placed at a distance $D$ is twice this angular spread multiplied by the distance because the central maximum spans symmetrically on both sides of the optical center. Combining these geometric factors yields the total central maxima width formula $\frac{2D\lambda}{a}$.

The refractive index of a prism is related to its refracting angle and the angle of minimum deviation through the prism formula $\mu = \frac{\sin\left(\frac{\delta_m+A}{2}\right)}{\sin\left(\frac{A}{2}\right)}$. Substituting the given refractive index condition $\mu = \cot\left(\frac{A}{2}\right)$, which expands to $\frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)}$, allows the denominator of the sine terms to cancel out on both sides. This simplifies the expression to $\cos\left(\frac{A}{2}\right) = \sin\left(\frac{\delta_m+A}{2}\right)$, which can be rewritten using complementary angle properties as $\sin\left(90^{\circ} - \frac{A}{2}\right) = \sin\left(\frac{\delta_m+A}{2}\right)$. Equating the arguments and solving for the angle of minimum deviation directly yields $\delta_m = 180^{\circ} - 2A$.

The focal length of a combination of lenses in contact is determined by summing the inverse focal lengths of the individual lens components using the lens maker's formula $\frac{1}{f} = (\mu - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$. Each glass plano-convex lens contributes a focal length where $\frac{1}{f_1} = (1.5 - 1)\left(\frac{1}{20}\right) = \frac{1}{40}$ and $\frac{1}{f_2} = (1.5 - 1)\left(\frac{1}{20}\right) = \frac{1}{40}$. The intervening oil forms a biconcave diverging lens with a refractive index of 1.7 and radii of curvature of 20 cm on both sides, yielding $\frac{1}{f_3} = (1.7 - 1)\left(-\frac{2}{20}\right) = -\frac{7}{100}$. Adding these reciprocal focal lengths together gives $\frac{1}{f} = \frac{1}{40} + \frac{1}{40} - \frac{7}{100}$, which simplifies to $-\frac{1}{50}$, resulting in a total focal length of -50 cm.

The intensity of light emerging from a slit is directly proportional to its width, and intensity varies as the square of the amplitude. From the given width ratio, the intensity ratio is established as $\frac{I_1}{I_2} = \frac{25}{1}$, leading to an amplitude ratio of $\frac{A_1}{A_2} = \frac{5}{1}$. Adding and subtracting these amplitudes gives the maximum and minimum values whose ratio is $\frac{3}{2}$, and squaring this result produces the final intensity ratio of $\frac{9}{4}$.

The intensity at any point in an interference pattern is determined by the phase difference between the interfering waves, expressed through the relation $I = I_{\max} \cos^2\left(\frac{\phi}{2}\right)$. A path difference of $\lambda$ yields the maximum intensity $K$, whereas a path difference of $\frac{\lambda}{4}$ introduces a corresponding phase difference of $\frac{\pi}{2}$. Substituting this phase angle into the intensity equation results in a final value of $\frac{K}{2}$.

The distance between the first dark fringes on either side of the central maximum in a single slit diffraction pattern equals the total width of the central maximum, which is expressed by the formula $\frac{2\lambda D}{d}$. Substituting the given values for wavelength, screen distance, and slit width into this expression evaluates to $2.4 \times 10^{-3}$ meters. Converting this length into millimeter units results in a final distance of $2.4$ mm.

When a light ray retraces its path after striking a silvered refracting surface, it must fall normally onto that surface, meaning the internal angle of refraction equals the prism angle $A$. Applying Snell's law at the first surface with an angle of incidence of $2A$ gives the refractive index as $\mu = \frac{\sin(2A)}{\sin(A)}$. Expanding the numerator using trigonometric identities simplifies this expression to $2 \cos A$.

The magnifying power of a compound microscope is inversely proportional to the focal length of its objective lens, whereas the magnifying power of an astronomical telescope is directly proportional to the objective focal length. Increasing the focal length of the objective lens therefore reduces the magnification of the microscope while simultaneously enhancing the magnification of the telescope.

The total converging power of the eye is the sum of meandering individual powers contributed by the cornea and the eye lens. Adding the corneal power of $40$ D and the minimum lens power of $20$ D yields a total optical power of $P = 60$ D for the eye system. Since focal length is the reciprocal of optical power, dividing $1$ by $60$ gives a focal length of $1/60$ m, which converts into $100/60$ cm to determine the distance to the retina as approximately $1.67$ cm.

Light from the sun travels through a much thicker layer of the Earth's atmosphere during sunrise and sunset than at noon. According to Rayleigh scattering, shorter wavelengths like blue and violet scatter much more easily in all directions by atmospheric molecules, while longer wavelengths like red remain largely unaffected and travel straight to our eyes. This differential depletion of the spectrum leaves the sun appearing distinctly reddish on the horizon.

The relationship between phase difference and path difference is given by the formula $\Delta \phi = \frac{2 \pi}{\lambda} \Delta x$. In single-slit diffraction, the second minimum corresponds to a total path difference of $2 \lambda$ between the edge rays. Substituting this path difference into the formula yields $\Delta \phi = \frac{2 \pi}{\lambda} \times 2 \lambda$, which evaluates to $4 \pi$.

The fringe width in Young's double slit experiment is governed by the relation $\beta = \frac{\lambda D}{d}$, where wavelength is constant, screen distance is represented by $D$, and slit separation is denoted by $d$. Doubling the distance to the screen scales the numerator by a factor of two, while halving the slit separation divides the denominator by two. Incorporating these proportional modifications into the expression changes the fringe width to four times its original value.

The focal length of a lens is determined by the lens maker's equation, $\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$. For the first plano-convex lens with refractive index $\mu_1$, the radii of curvature are $R_1 = \infty$ and $R_2 = -R$, yielding a focal length given by $\frac{1}{f_1} = \frac{\mu_1 - 1}{R}$. For the second plano-concave lens with refractive index $\mu_2$, the radii are $R_1 = -R$ and $R_2 = \infty$, giving $\frac{1}{f_2} = \frac{1 - \mu_2}{R}$. Combining these two lenses in contact means adding their optical powers, so substituting the individual focal lengths into $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$ simplifies the expression to $\frac{1}{F} = \frac{\mu_1 - \mu_2}{R}$, which directly yields the combined focal length.

Interference fringes of different wavelengths coincide on a screen where the positions of their bright fringes match, expressed by equating their fringe position formulas $\frac{n_1 \lambda_1 D}{d} = \frac{n_2 \lambda_2 D}{d}$. Cancelling the common distance $D$ and slit separation $d$ leads to the condition $n_1 \lambda_1 = n_2 \lambda_2$, which gives the ratio of integers $n_1 / n_2 = \lambda_2 / \lambda_1$. Substituting the given values $10000$ Å and $12000$ Å yields the fraction $5 / 6$, meaning the fifth bright fringe of the first wavelength coincides with the sixth of the second. Using $n_1 = 5$ alongside $D = 2$ m, $d = 2$ mm, and $\lambda_1 = 12000$ Å in the position formula results in a minimum distance of $6$ mm.

The successive reflection of light between two inclined mirrors is governed by the laws of reflection and the geometric properties of triangles. Given that the final reflected ray is parallel to the first mirror, the geometry dictates that the angle of incidence on the second surface creates a complementary relationship at the vertex. Using the triangle angle sum property with the mirror inclination, the angle adjacent to the second reflection leads to a simple linear relation for the first interface, expressed as $\theta + 40^{\circ} = 90^{\circ}$. Solving this equation directly yields $\theta = 50^{\circ}$.

The angular width of a central diffraction maximum is directly proportional to the wavelength of the incident particles and inversely proportional to the slit dimension, represented by $\omega = \frac{2\lambda}{d}$. De Broglie's relation $\lambda = \frac{h}{m v}$ establishes that particle wavelength decreases as velocity increases because momentum becomes larger. Consequently, when the electron speed is raised, their shorter wavelengths produce a narrower diffraction pattern, leading to a decrease in the angular width of the central maximum.

The deviation produced by a prism is related to the angles of incidence and emergence through the equation $\delta = i + e - A$. Since the ray emerges normally from the opposite face, the angle of emergence is zero. Substituting this condition simplifies the deviation to $\delta = i - A$. For a small-angle prism, the deviation is also expressed as $\delta = (\mu - 1)A$, and equating these two expressions directly yields the angle of incidence as $\mu A$.

The mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ governs the image formation for each point along the optical axis, with all distances adhering to standard sign conventions. For the closer end of the rod positioned at an object distance of 20 cm, substituting a focal length of -10 cm and an object distance of -20 cm into the formula yields an image position at 20 cm in front of the mirror. Applying the same mirror equation to the farther end of the rod located at an object distance of 30 cm results in an image position at 15 cm. The length of the image is then obtained by calculating the difference between these two image distances, yielding a final length of 5 cm.