Ray Optics And Optical Instruments – JEE Main PhysicsPractice Questions & PYQs

Finding the image length for an extended object placed along the principal axis involves locating the positions of the two endpoints using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$. For a concave mirror with a focal length of -10 cm, we calculate the image location for the first endpoint positioned at -20 cm using the rearranged expression $v = \frac{u \times f}{u - f}$, which results in an image distance of -20 cm.

Performing the same calculation for the second endpoint at -30 cm yields an image position of -15 cm. The image formed by the rod spans the distance between these two calculated points, so the length is determined by the absolute difference between -20 cm and -15 cm, resulting in a final length of 5 cm.

When the reflected and refracted rays are perpendicular, the angle of incidence is known as the Brewster's angle, denoted by $\theta_p$, which satisfies the relationship $\tan \theta_p = \frac{n_2}{n_1}$, where $n_1$ and $n_2$ are the refractive indices of the incident and refracting media, respectively. Inserting the given values $n_1 = 2$ and $n_2 = 2\sqrt{3}$ into this formula yields $\tan \theta_p = \frac{2\sqrt{3}}{2} = \sqrt{3}$. Consequently, the angle of incidence is $\theta_p = \arctan(\sqrt{3}) = 60^\circ$.

When the incident ray is parallel to the base of the prism, the path of the light becomes symmetric, ensuring that the angles of refraction at the first and second surfaces are equal, defined as $r_1 = r_2 = \theta / 2$. Applying Snell's Law at the second surface, where the emergent ray grazes the prism surface at an angle of $e = 90^{\circ}$, gives the equation $\mu \sin(r_2) = \sin(90^{\circ}) = 1$. Substituting $r_2 = \theta / 2$ into this expression yields $\sin(\theta / 2) = 1 / \mu$. Using the refractive index $\mu = 2$, this simplifies to $\sin(\theta / 2) = 1 / 2$, which corresponds to an angle of $\theta / 2 = 30^{\circ}$, resulting in a prism angle of $\theta = 60^{\circ}$.

The refractive index of a prism is determined by the relationship $\mu = \frac{\sin\left(\frac{A+\delta_{\min}}{2}\right)}{\sin\left(\frac{A}{2}\right)}$, where $A$ is the prism angle and $\delta_{\min}$ is the angle of minimum deviation. In an equilateral prism, the angle is $A = 60^\circ$, and with the angle of minimum deviation given as half of that value, we have $\delta_{\min} = 30^\circ$. Substituting these values into the governing equation yields $\mu = \frac{\sin\left(\frac{60^\circ + 30^\circ}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)} = \frac{\sin(45^\circ)}{\sin(30^\circ)}$. Evaluating the trigonometric functions with $\sin(45^\circ) = \frac{1}{\sqrt{2}}$ and $\sin(30^\circ) = \frac{1}{2}$ results in the final calculation of $\mu = \sqrt{2}$.

The critical angle $\theta_c$ for total internal reflection at an interface is determined by the ratio of the speeds of light in the two media, where $\sin \theta_c = \frac{v_1}{v_2}$. Substituting the given values $v_1 = 2.4 \times 10^8 \text{ m/s}$ and $v_2 = 2.7 \times 10^8 \text{ m/s}$ results in $\sin \theta_c = \frac{2.4}{2.7} = \frac{8}{9}$.

Using the identity $\tan \theta = \frac{\sin \theta}{\sqrt{1 - \sin^2 \theta}}$, the angle can be expressed as:

$\tan \theta_c = \frac{8/9}{\sqrt{1 - (8/9)^2}} = \frac{8/9}{\sqrt{1 - 64/81}} = \frac{8/9}{\sqrt{17/81}} = \frac{8}{\sqrt{17}}$

The resulting critical angle is $\tan^{-1}\left(\frac{8}{\sqrt{17}}\right)$.

For total internal reflection to occur at the back exit surface of the prism, the angle of incidence at that surface ($r_2$) must be greater than or equal to the critical angle $C$, where the condition is defined by $\sin C = \frac{\mu_f}{\mu_p}$. Given the prism geometry $r_1 + r_2 = A$, where $A = 75^{\circ}$, the requirement $r_2 \geq C$ can be expressed as $r_1 \leq A - C$.

Substituting the given values, the critical angle is $\sin C = \frac{1.5}{3\sqrt{3}} = \frac{1}{2\sqrt{3}} \approx 0.288$, which corresponds to $C \approx 16.7^{\circ}$. Consequently, the condition for total internal reflection becomes $r_1 \leq 75^{\circ} - 16.7^{\circ} = 58.3^{\circ}$. Applying Snell's Law at the first surface, $\sin i = \mu_p \sin r_1$, we see that a smaller incident angle $i$ corresponds to a smaller refraction angle $r_1$, which in turn ensures a larger $r_2$ that easily exceeds the critical angle. An incident angle $i < 15^{\circ}$ results in a very small $r_1$, keeping $r_2$ well above the critical angle and guaranteeing total internal reflection at the back exit surface.

A concave mirror produces a magnification of 2 in two distinct scenarios: one resulting in a real, inverted image ($m = -2$) and the other in a virtual, erect image ($m = +2$). Since the magnification formula is $m = -\frac{v}{u}$, for the real image case where $m = -2$, we have $v = 2u$. Applying the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ with the appropriate sign convention, where $u$ is the object distance and $f = -10$ cm, we substitute the values to get $\frac{1}{-2u} + \frac{1}{-u} = \frac{1}{-10}$, which simplifies to $\frac{3}{2u} = \frac{1}{10}$, yielding a first object position at $u_1 = 15$ cm.

For the virtual image case where $m = +2$, the relationship becomes $v = -2u$. By applying the mirror formula again with the sign convention where the virtual image distance is positive and the object distance is negative, we use $\frac{1}{2u} + \frac{1}{-u} = \frac{1}{-10}$. This simplifies to $-\frac{1}{2u} = -\frac{1}{10}$, which gives a second object position at $u_2 = 5$ cm. The separation between these two positions is simply the difference between the two calculated object distances, which is $15\text{ cm} - 5\text{ cm} = 10$ cm.

The refraction of light rays at a spherical interface is governed by the paraxial approximation formula, which defines the relationship between the object distance, the image distance, and the optical properties of the two media. This principle allows us to determine the image position by utilizing the refractive indices $\mu_1 = 1$ and $\mu_2 = 1.4$ for a source positioned at $u = -4R$.

Substituting these parameters into the general refraction expression $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$ leads to $\frac{1.4}{v} + \frac{1}{4R} = \frac{0.4}{R}$. Isolating the term $\frac{1.4}{v}$ gives $\frac{0.4}{R} - \frac{0.25}{R} = \frac{0.15}{R}$, which simplifies to an image distance of $v = \frac{1.4}{0.15}R = \frac{28}{3}R$. The magnitude of the transverse magnification is then determined by the ratio $m = \left| \frac{\mu_1 v}{\mu_2 u} \right|$. Applying the calculated image distance and given object position yields a magnitude of $\left| \frac{1 \times (28R/3)}{1.4 \times (-4R)} \right| = \left| \frac{28}{3 \times -5.6} \right| = \left| \frac{28}{-16.8} \right| = 1.66$.

A thin prism is defined by the relationship between the angle of minimum deviation, $\delta_m$, the refractive index of the material, $\mu$, and the prism angle, $A$, which is expressed as $\delta_m = (\mu - 1)A$. With a glass refractive index of 1.5, the angle of minimum deviation becomes $\delta_m = (1.5 - 1)A$, simplifying to $\delta_m = \frac{A}{2}$. During minimum deviation, the incident angle $i$ is determined by the relation $i = \frac{A + \delta_m}{2}$. Substituting the value of $\delta_m$ into this equation gives $i = \frac{A + \frac{A}{2}}{2} = \frac{3A}{4}$. The final ratio of the incident angle to the angle of minimum deviation is calculated as $\frac{i}{\delta_m} = \frac{\frac{3A}{4}}{\frac{A}{2}} = \frac{3}{2}$.

The total magnification $M$ of a compound microscope in normal adjustment, where the final image is formed at infinity, is calculated using the formula $M = \left( \frac{L}{f_o} \right) \left( \frac{D}{f_e} \right)$, where $L$ is the tube length, $f_o$ and $f_e$ are the focal lengths of the objective and eyepiece respectively, and $D = 25 \text{ cm}$ is the least distance of distinct vision. Substituting the given values $L = 10 \text{ cm}$, $f_o = 2 \text{ cm}$, and $f_e = 5 \text{ cm}$ into the expression yields $M = \left( \frac{10}{2} \right) \left( \frac{25}{5} \right) = 5 \times 5 = 25$. Since the magnification is represented as $(5)^k$, we equate $25 = 5^2$, which shows that the value of $k$ is 2.

Snell's law can be expressed in vector form by equating the cross products of the unit vectors of the incident and refracted rays with the normal vector at the interface. Assuming the interface lies in the horizontal plane with the normal along the $\hat{j}$ direction, the physical principle states that the cross product of the incident unit vector with the normal is equal to the product of the relative refractive index and the cross product of the refracted unit vector with the normal. Here, the refractive index ratio is $\mu = 1.5$, and we first define the unit vectors by normalizing the given rays, resulting in $\hat{u}_{AO} = \frac{2\hat{i} - 3\hat{j}}{\sqrt{13}}$ and $\hat{u}_{OB} = \frac{C\hat{i} - 4\hat{j}}{\sqrt{C^{2+}16}}$.

Taking the cross product of each unit vector with the normal vector $\hat{j}$ yields $\frac{2\hat{k}}{\sqrt{13}}$ for the incident ray and $\frac{C\hat{k}}{\sqrt{C^{2+}16}}$ for the refracted ray. Substituting these results into the vector form of Snell's law gives the equation $\frac{2}{\sqrt{13}} = 1.5 \left( \frac{C}{\sqrt{C^{2+}16}} \right)$, which simplifies to $\frac{2}{\sqrt{13}} = \frac{3C}{2\sqrt{C^{2+}16}}$. Squaring both sides of this equation produces $\frac{4}{13} = \frac{9C^2}{4(C^{2+}16)}$, which rearranges to $16(C^{2+}16) = 117C^2$ after cross-multiplication. Expanding this leads to $16C^2 + 256 = 117C^2$, and isolating the variable results in $101C^2 = 256$. Solving for $C$ gives $C^2 = \frac{256}{101} \approx 2.53$, and taking the square root yields $C \approx 1.59$, which is approximately $1.6$.

The magnification $M$ of a compound microscope in normal adjustment is calculated using the formula $M = \left( \frac{L}{f_o} \right) \left( \frac{D}{f_e} \right)$, where $L$ is the tube length, $f_o$ and $f_e$ are the focal lengths of the objective and eyepiece respectively, and $D$ is the least distance of distinct vision, which is standardly taken as $25 \text{ cm}$.

Substituting the given values $L = 32 \text{ cm}$, $f_o = 2 \text{ cm}$, $f_e = 4 \text{ cm}$, and $D = 25 \text{ cm}$ into the equation results in $M = \left( \frac{32}{2} \right) \left( \frac{25}{4} \right)$. Simplifying this expression gives $M = 16 \times 6.25$, which yields a total magnification of $100$.

When light travels through a single spherical surface separating two media with different refractive indices, its behavior is governed by the refraction formula $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$. By applying the given values where $\mu_1 = 1$ represents the refractive index of air, $\mu_2 = 1.54$ is the refractive index of the lens material, $u = -40$ cm is the object distance, and $R = -20$ cm is the radius of curvature, we can write the equation as $\frac{1.54}{v} - \frac{1}{-40} = \frac{1.54 - 1}{-20}$. Rearranging the terms to isolate the variable for the image position leads to $\frac{1.54}{v} = - \frac{1}{40} - \frac{0.54}{20}$, which further simplifies to $\frac{1.54}{v} = -\frac{2.08}{40}$, resulting in an image distance of approximately $v = -29.61$ cm.

To determine the final image height, we utilize the lateral magnification relation for refraction at a spherical surface, expressed as $\frac{h_i}{h_o} = \frac{\mu_1 v}{\mu_2 u}$. Substituting the known values along with the object height of $h_o = 2$ cm, we compute $h_i = 2 \times \frac{1 \times (-29.61)}{1.54 \times (-40)}$. This calculation provides a magnification factor of approximately 0.48, which, when multiplied by the original height of the object, yields an image height of 0.96 cm, rounding to 1 cm.

The Lens Maker's Formula provides the standard method to calculate the focal length $f$ of a lens based on its refractive index and surface curvatures. Given that the lens is convex with equivalent radii on both sides, we apply the sign convention where the first surface has a positive radius $R$ and the second has a negative radius $-R$. Substituting the refractive index of 1.4 into the formula results in $\frac{1}{f} = (1.4 - 1) \left( \frac{1}{R} - \frac{1}{-R} \right) = \frac{0.8}{R}$. Rearranging this expression to isolate the ratio of focal length to the radius of curvature gives $\frac{f}{R} = \frac{1}{0.8} = 1.25$.

For a prism at the condition of minimum deviation, the angle of refraction $r$ is equal at both faces and is given by $r = \frac{A}{2}$, where $A$ is the prism angle. The condition for the critical angle at the exit surface is defined by Snell's law where the angle of incidence from the prism to the coating equals the critical angle, resulting in $\sin(r_2) = \frac{n_2}{n}$, where $n_2$ is the refractive index of the coating and $n$ is the refractive index of the prism. Substituting $r_2 = \frac{A}{2}$ and $n_2 = \frac{n}{2}$ into this relationship yields $\sin\left(\frac{A}{2}\right) = \frac{n/2}{n} = \frac{1}{2}$. Solving for the prism angle gives $\frac{A}{2} = 30^{\circ}$, which results in $A = 60^{\circ}$.

The system relies on the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$, where the image from lens P serves as the object for lens Q. Substituting $u = -15$ cm and $f = 10$ cm for lens P into the equation $\frac{1}{v} - \frac{1}{-15} = \frac{1}{10}$ places the first image $v = 30$ cm to the right of lens P, which acts as a virtual object for lens Q at $u = +15$ cm from its center. Using $f = 15$ cm for lens Q, the equation $\frac{1}{v'} - \frac{1}{15} = \frac{1}{15}$ results in $v' = 7.5$ cm, while the total magnification $m = (\frac{30}{-15}) \times (\frac{7.5}{15}) = -1$ confirms the final image is real, equal in size to the object, and formed 7.5 cm to the right of lens Q.

Total internal reflection occurs when a light ray traveling within a denser medium encounters a boundary with a less dense medium at an angle of incidence equal to or exceeding the critical angle. In an equilateral prism, a ray entering perpendicular to one face strikes the next face at an incidence angle of $60^{\circ}$. Applying Snell's law at this boundary leads to the condition:

$1.6 \sin 60^{\circ} = n_2 \sin 90^{\circ}$

Solving for the refractive index $n_2$ required for the refracted angle to reach $90^{\circ}$ results in $n_2 = 1.6 \times \frac{\sqrt{3}}{2}$, which simplifies to $\frac{4\sqrt{3}}{5}$.

For a thin convex lens to produce images of the same size at two different object positions $d_1$ and $d_2$, the images must have opposite characteristics, meaning one is real and inverted while the other is virtual and erect, implying their magnifications are $m$ and $-m$ respectively. Given the magnification formula $m = \frac{f}{f+u}$ where $u = -d$ (using the sign convention for object distance), the condition of equal image sizes $|m_1| = |m_2|$ leads to the relationship $\frac{f}{f-d_1} = -\frac{f}{f-d_2}$. Substituting the given object distances $d_1 = 8$ cm and $d_2 = 24$ cm, the equation becomes $\frac{f}{f-8} = -\frac{f}{f-24}$. Canceling $f$ from both sides and cross-multiplying, we get $f-24 = -(f-8)$, which simplifies to $2f = 32$, resulting in a focal length of $f = 16$ cm.

The thin lens formula is given by $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$, where the focal length $f = \frac{1}{P} = \frac{1}{5} \text{ m} = 20 \text{ cm}$. Applying the sign convention for a convex lens, where the object distance $u$ is treated as a negative value, the equation becomes:
$\frac{1}{v} + \frac{1}{u} = \frac{1}{20} = 0.05$
Substituting the provided values for each person, we obtain $\frac{1}{96} + \frac{1}{25} \approx 0.0504$ for $P_1$, $\frac{1}{62} + \frac{1}{30} \approx 0.0494$ for $P_2$, $\frac{1}{37} + \frac{1}{35} \approx 0.0556$ for $P_3$, $\frac{1}{35} + \frac{1}{45} \approx 0.0508$ for $P_4$, and $\frac{1}{32} + \frac{1}{50} = 0.05125$ for $P_5$. Among these calculations, the value for $P_3$ deviates most significantly from the required constant of $0.05$, identifying it as the incorrect measurement.

A biconvex lens silvered on its rear surface effectively behaves as a single concave mirror because incoming light travels through the lens, reflects off the silvered back, and refracts through the lens once more. The focal length of the lens component is $f_L = 20\text{ cm}$, while the silvered surface serves as a mirror with a focal length of $f_m = -10\text{ cm}$. Combining these optical properties, the equivalent focal length $f$ of the system follows the relationship:

$\frac{1}{f} = \frac{1}{f_m} - \frac{2}{f_L}$

Substituting the specific values gives $\frac{1}{f} = -\frac{1}{10} - \frac{2}{20} = -\frac{1}{5}$, which results in an effective focal length of $f = -5\text{ cm}$. Because the light must retrace its own path to create an image at the same position as the object, the object must be situated at the center of curvature of this equivalent mirror, corresponding to a distance of $u = 2f = 10\text{ cm}$.