Optics – NEET UG PhysicsPractice Questions & PYQs

When parallel rays from infinity pass through a convex lens, they converge at its focal point situated at a distance $f_2$ from the lens. For these rays to retrace their path and return to infinity after reflecting from the concave mirror, they must strike the mirror's reflecting surface normally. This requirement implies that the focal point of the lens must coincide with the center of curvature of the concave mirror. Since the center of curvature lies at a distance of $2f_1$ from the pole of the mirror, the total distance $d$ between the two optical elements equals $2f_1 + f_2$.

A lens loses its refracting property and acts as a flat plane sheet when its focal length becomes infinite, meaning light passes through without any bending. Based on the lens maker formula $\frac{1}{f} = \left(\frac{\mu_g}{\mu_L} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$, this optical behavior occurs when the refractive index ratio between the lens and the surrounding medium equals one. Therefore, the liquid surrounding the glass must possess a refractive index exactly equal to that of the glass itself.

The refractive index of a prism is given by the expression $\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$, where $A$ is the refracting angle and $\delta_m$ is the angle of minimum deviation. Applying the given condition that the minimum deviation equals the refracting angle transforms the expression into $\mu = 2\cos\left(\frac{A}{2}\right)$ after trigonometric simplification. As the refracting angle ranges across its geometric limits, the resulting refractive index is restricted to values between 2 and $\sqrt{2}$.

The magnifying power for a telescope in normal adjustment is given by the formula $m = \frac{f_o}{f_e}$, where $f_o$ and $f_e$ represent the focal lengths of the objective and eyepiece lenses. Given a magnification of 9, the focal length of the objective is nine \times that of the eyepiece, meaning $f_o = 9f_e$. The total distance between the two lenses when adjusted for parallel rays is the \sum of their focal lengths, expressed as $f_o + f_e = 20$. Substituting the first relation into this \sum yields $10f_e = 20$, leading to an eyepiece focal length of 2 cm and an objective focal length of 18 cm.

Dispersion without deviation requires the net deviation produced by the prism combination to be zero, meaning the sum of the individual deviations must satisfy $\delta_1 + \delta_2 = 0$. Expressing deviation in terms of prism angle and refractive index transforms this condition into $(\mu_1 - 1)A_1 + (\mu_2 - 1)A_2 = 0$. Rearranging this equation isolates the required angle of the second prism as $A_2 = -\frac{(\mu_1 - 1)A_1}{\mu_2 - 1}$. Substituting the given values yields $A_2 = -\frac{(1.5 - 1)15^{\circ}}{1.75 - 1}$, which calculates to $-10^{\circ}$, where the negative sign signifies that the second prism is oriented in the opposite direction.

A converging beam encountering a diverging lens has a virtual object position corresponding to the point where the rays would have converged without the lens. Removing the lens shifts this convergence point $5$ cm closer, establishing that the object distance is $10$ cm while the image distance after refraction is $15$ cm. Substituting these values into the standard thin lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ provides $\frac{1}{15} - \frac{1}{10} = \frac{1}{f}$. Solving this algebraic expression directly results in a focal length of $-30$ cm, characteristic of a diverging lens.

Total internal reflection happens when light travels from a denser to a rarer medium and strikes the interface at an angle greater than the critical angle, causing it to reflect entirely back inside. Optical fibers, mirages, and the brilliance of diamonds all rely on this trapped reflection mechanism to function or create their visual effects. In contrast, viewing the bottom of a pond involves light crossing from water into the air above, where it bends due to refraction rather than reflecting internally. This bending alters our line of sight, making the water's bottom appear closer to the surface than it actually is.

The focal length is found via the lens maker's equation $\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$, giving $20$ cm for standard glass. Substituting this focal length and the object distance into the thin lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ places the image at a distance of $60$ cm. The transverse magnification is expressed by $\frac{v}{u}$, yielding a value of $-2$. Multiplying this magnification ratio by the initial object height of $2$ cm produces an image height of $-4$ cm, indicating a real and inverted image.

Total internal reflection occurs when the angle of incidence exceeds the critical angle of the medium, establishing the mathematical condition $\mu > \frac{1}{\sin i}$. Substituting the given incidence angle into this inequality gives $\mu > \frac{1}{\sin 45^{\circ}}$, which simplifies to $\mu > \sqrt{2}$. Since the square root of two is approximately $1.414$, the refractive index of the medium must be strictly greater than this threshold. Among the given choices, only a refractive index of $1.50$ satisfies this requirement.

For a prism, the sum of the angles of refraction at the first and second surfaces always equals the refracting angle of the prism itself. When the prism is adjusted to the position of minimum deviation, the ray passes symmetrically through the interior, causing the angles of refraction at both faces to become equal. Given that the total prism angle is 60°, this symmetric condition splits the value evenly between the two refracting surfaces, resulting in a first-surface refraction angle of 30°.

The absolute refractive index of any medium is determined by dividing the speed of light in a vacuum by the speed of light within that specific medium. Applying this relation yields a refractive index of 2 for the first medium and 3/2 for the second medium. Total internal reflection takes place when a ray attempts to pass into a rarer medium at an incidence angle greater than or equal to the critical angle, satisfying the condition $\sin i \geq \mu_2 / \mu_1$ which evaluates to $\sin^{-1}(3/4)$.

The focal length of a lens depends strictly on its refractive index and the radii of curvature of its surfaces, remaining entirely unaffected when a part of the aperture is blocked. However, the brightness or intensity of the formed image is directly proportional to the exposed surface area of the lens available for collecting light. The initial area of the full aperture of diameter $d$ is given by $A_{1} = \frac{\pi d^{2}}{4}$, and covering the central region reduces the effective area to $A_{2} = \frac{3 \pi d^{2}}{16}$. Taking the ratio of the new area to the initial area reveals that the intensity drops to $\frac{3}{4}$ of its original value $I$, while the focal length stays at $f$.

When multiple thin lenses are placed in direct coaxial contact, the reciprocal of the equivalent focal length of the combined system is equal to the sum of the reciprocals of the individual focal lengths. Combining these fractions finds a common denominator, expressing the net focal length in terms of the product over the sum of the individual lenses. Because optical power is mathematically defined as the reciprocal of the focal length, taking this inverse directly yields the combined power formula $\frac{f_1 + f_2}{f_1 f_2}$.

When an astronomical object like the sun is located at an extremely large distance from a lens, light rays arrive essentially parallel and form an image at the focal plane of the lens. The linear magnification produced by the optical system relates the size of the image to the size of the object through the ratio of the image distance to the object distance. Substituting the focal length of 10 cm for the image distance and the sun's distance from earth into this proportion yields a final image diameter of 9.2 x 10^-4 m.

When a light ray travels from a denser medium and moves along the boundary surface, the angle of incidence inside the liquid reaches the critical angle. The fundamental relation connecting this critical angle to wave speeds is given by the formula $\sin C = \frac{v}{c}$, where $v$ is the speed of light in the liquid and $c$ is the speed of light in air. Given that the sine of the critical angle is $3/5$, the velocity can be determined through the equation $v = 3 \times 10^{8} \times \frac{3}{5}$ which calculates out to 1.8 × 10^8 m/s.

The refractive index of a medium is computed by dividing the speed of light in a vacuum by the actual speed of light traveling within that material. The speed of light inside the medium can be determined by multiplying its wave frequency by its corresponding wavelength in meters. Performing this multiplication for the given frequency and wavelength results in a medium speed of 10^8 m/s, which when compared against the vacuum speed of light produces a refractive index of 3.

The apparent shift of an object viewed through a refracting medium is determined by the relation $\text{Apparent Depth} = \frac{\text{Real Depth}}{\mu}$. Dividing the real depth of 3 cm by the refractive index of 1.5 yields an apparent depth of 2 cm. Because the image appears raised by the difference between these values, which equals 1 cm, the microscope must be moved upward by 1 cm to restore sharp focus.

The net optical power of thin lenses placed in direct contact is given by the algebraic sum of their individual powers, expressed as $P = P_1 + P_2$, where each power is calculated as 100 divided by the focal length in centimeters. For the convex lens, the focal length is positive, yielding a power of 4 D, while the concave lens has an equal and opposite negative focal length, yielding -4 D. Combining these two values results in a net power of zero for the system.

The angular resolution of a telescope limited by diffraction through a circular aperture is governed by the formula $\theta = 1.22 \frac{\lambda}{d}$. Substituting the wavelength of 5000 Å (equivalent to $5000 \times 10^{-8}$ cm) and the aperture diameter of 10 cm into the expression $\theta = 1.22 \frac{5000 \times 10^{-8}}{10}$ evaluates to $6.1 \times 10^{-6}$ rad. This places the order of magnitude of the resolution at $10^{-6}$ rad.

The minimum resolvable linear distance between two distant objects is determined by multiplying the angular resolution ofয়ে the telescope by the viewing distance, expressed as $x = \left( 1.22 \frac{\lambda}{D} \right) d$. Substituting the wavelength of 5000 Å ($5000 \times 10^{-8}$ cm), the objective diameter of 10 cm, and the distance of 1 km ($10^5$ cm) into $x = \left( 1.22 \times \frac{5000 \times 10^{-8}}{10} \right) \times 10^5$ yields approximately 6.1 mm. This computed value falls on the order of 5 mm.