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

The Lens Maker's Formula relates a lens's focal length to its refractive index and the curvature of its surfaces, expressed as $\frac{1}{f} = (\mu - 1)(\frac{1}{R_1} - \frac{1}{R_2})$. Applying the sign convention for a convex lens where the first surface has a radius of 15 cm and the second has a radius of -30 cm, we substitute the known focal length $f = 20$ cm into the equation to get $\frac{1}{20} = (\mu - 1)(\frac{1}{15} - \frac{1}{-30})$. Evaluating the bracketed terms gives $\frac{1}{15} + \frac{1}{30} = \frac{3}{30}$, which simplifies to $\frac{1}{10}$, so the equation reduces to $\frac{1}{20} = (\mu - 1)(\frac{1}{10})$. Solving for $\mu$ yields $\mu - 1 = 0.5$, which determines the refractive index of the material to be 1.5.

When light travels from an object through multiple layers of immiscible liquids to an observer, the apparent depth of the object is determined by the sum of the contributions from each individual liquid layer. Each layer shifts the position of the image because its refractive index causes the light rays to bend, effectively reducing the perceived depth of that section. The total apparent depth is calculated using the relation $h_{app} = \frac{h_1}{\mu_1} + \frac{h_2}{\mu_2}$ where $h_1$ and $h_2$ are the heights of the individual layers and $\mu_1$ and $\mu_2$ are their respective refractive indices. With both layers having a height of 6 cm and refractive indices of 8/5 and 3/2, the total apparent depth becomes 6 / (8/5) + 6 / (3/2). Solving these terms, the first part simplifies to 15/4 and the second part simplifies to 4, which combine to a total apparent depth of 31/4 cm. As the problem defines the apparent depth as $\frac{\alpha}{4}$ cm, equating this to our result confirms that $\alpha$ is 31.

Lateral displacement occurs when a light ray traverses a parallel-sided slab, and its magnitude is determined by the slab's thickness and the refraction angles. Because the incident angle matches the critical angle for the glass-air interface, we use the definition $\sin i = \frac{1}{\mu}$. Given a refractive index of $\sqrt{2}$, $\sin i$ becomes $\frac{1}{\sqrt{2}}$, which fixes the angle of incidence at $45^{\circ}$. Snell's Law at the first surface, $1 \cdot \sin 45^{\circ} = \sqrt{2} \sin r$, then simplifies to $\sin r = 0.5$, confirming an angle of refraction of $30^{\circ}$.

The lateral shift is calculated using the relation $\Delta = \frac{t \sin (i - r)}{\cos r}$. Plugging in the slab thickness $4\sqrt{3}$ cm along with the calculated angles, we have $\Delta = \frac{4\sqrt{3} \sin 15^{\circ}}{\cos 30^{\circ}}$. Using the provided value $\sin 15^{\circ} = 0.25$ and the trigonometric value $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$, the equation becomes $\Delta = \frac{4\sqrt{3} \cdot 0.25}{\frac{\sqrt{3}}{2}}$. Simplifying this expression involves canceling $\sqrt{3}$ from the numerator and denominator, which leaves $4 \cdot 0.25 \cdot 2$, resulting in a final displacement of 2 cm.

A concave lens is fundamentally a diverging lens, meaning it causes incoming light rays to spread out after refraction rather than converging at a point. By applying the lens equation $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ with the standard sign convention where the focal length $f$ is negative, placing an object at the centre of curvature $u = -2f$ results in the calculation $\frac{1}{v} - \frac{1}{-2f} = \frac{1}{-f}$, which yields $v = -\frac{2}{3}f$. This confirms that the image forms on the same side as the object, rather than at the centre of curvature on the opposite side, rendering the first statement incorrect.

For any real object positioned in front of a concave lens, the refracted light rays diverge and appear to originate from a point on the same side as the object. This consistent optical behavior ensures that the resulting image is always virtual, erect, and diminished in size, which validates the general assertion made in the second statement.

For a real image formed by a convex lens, the magnification is given by $m = v/u = -2$, which establishes the relationship $v = -2u$ between the image and object distances. The total separation between the object and its real image is expressed as $v - u = 45$ cm, allowing us to substitute the magnification relation to get $-2u - u = 45$ cm, which simplifies to $-3u = 45$ cm, resulting in an object position of $u = -15$ cm and an image position of $v = 30$ cm. Using these values in the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ leads to $\frac{1}{f} = \frac{1}{30} - \frac{1}{-15}$, which further simplifies to $\frac{1}{f} = \frac{1}{30} + \frac{2}{30} = \frac{3}{30}$. This calculation confirms that the focal length of the lens is 10 cm.

The position of the final image is determined by applying the standard lens formula $v = \frac{uf}{u+f}$ iteratively for each optical element in the system. Starting with the first lens, which has a focal length of 10 cm and an object distance of -30 cm, the light rays form an image at a distance of 15 cm. This location effectively functions as the object position for the second lens in the series.

For the second lens, which has a focal length of -10 cm, the incoming rays act as an object at a distance of 10 cm. Applying the same formula results in $v = \infty$, indicating that the light rays emerge parallel to the principal axis after refraction. Since these collimated rays are incident upon the third lens with a focal length of 30 cm, they converge at the lens's focal point, placing the final image at 30 cm to the right of the third lens.

The relationship between the refractive index $\mu$ of a prism, its apex angle $A$, and the angle of minimum deviation $\delta_m$ is defined by the formula $\mu = \frac{\sin\left(\frac{A+\delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$. Given that $\mu = \cot\left(\frac{A}{2}\right)$, we equate these expressions to get $\frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin\left(\frac{A+\delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$. Simplifying by removing the common denominator, we arrive at $\cos\left(\frac{A}{2}\right) = \sin\left(\frac{A+\delta_m}{2}\right)$. Applying the trigonometric identity $\cos(\theta) = \sin(90^{\circ} - \theta)$, we rewrite the left side as $\sin\left(90^{\circ} - \frac{A}{2}\right) = \sin\left(\frac{A+\delta_m}{2}\right)$. Equating the arguments leads to $90^{\circ} - \frac{A}{2} = \frac{A+\delta_m}{2}$, which rearranges to $180^{\circ} - A = A + \delta_m$, resulting in $\delta_m = 180^{\circ} - 2A$.

When object distance $x$ and image distance $y$ are measured relative to the focal points of a convex lens, the system follows Newton's form of the lens equation, which is expressed as $xy = f^2$. This geometric relationship indicates that the product of these two distances equals the square of the focal length. Looking at the provided plot, we observe that the measurements for $x$ and $y$ intersect where both values are 20 cm. Substituting these values into the governing equation yields $f^2 = 20 \times 20$, which confirms that the focal length is 20 cm.

When thin lenses are placed in contact along the same principal axis, their combined power is the algebraic sum of the individual powers of each lens. Since there are 5 identical convex lenses, the total power is 5 times the power of a single lens, which is expressed by the equation $P_{\text{eq}} = 5P$. Given that the total equivalent power is 25 D, we substitute this value into the equation to get $5P = 25$. Solving for the power of one lens yields $P = 5$ D. Because power is the reciprocal of the focal length in meters, expressed as $P = \frac{1}{f}$, we find the focal length by calculating $f = \frac{1}{5}$ meters. Converting this result into centimeters, the focal length is 20 cm.

A spherometer is a precision mechanical instrument designed to determine the radius of curvature of spherical surfaces or the thickness of small objects by measuring the sagitta, $h$, through the vertical displacement of its central screw. Because its operation relies on physical contact to define the geometry of a surface, it is inherently suited to measure the radius of curvature for both concave and convex surfaces, as well as the linear thickness of thin plates using the relationship $R = \frac{l^2}{6h} + \frac{h}{2}$. Conversely, the specific rotation of liquids describes the optical ability of a substance to rotate the plane of polarized light, a property that necessitates the use of a polarimeter rather than a mechanical contact-based device.

The magnitude of the photoelectric current depends primarily on the total radiant energy captured and focused by the lens, which is determined by the lens diameter rather than its focal length. Since the diameter remains unchanged while replacing the 40 cm focal length lens with a 20 cm one, the lens continues to intercept the same amount of light energy from the extended source. Because the total energy hitting the cell stays constant, the photoelectric current remains equal to $I$.

The critical angle represents the angle of incidence when light traveling from a denser medium to a rarer medium is refracted at an angle of $90^\circ$. The relationship between the critical angle $\theta_c$ and the refractive indices of the denser medium, $\mu_1$, and the rarer medium, $\mu_2$, is defined as:

$\sin \theta_c = \frac{\mu_2}{\mu_1}$

Substituting the given critical angle of $45^\circ$ into this expression provides $\sin 45^\circ = \frac{\mu_2}{\mu_1}$. Given that $\sin 45^\circ = \frac{1}{\sqrt{2}}$, we have $\frac{1}{\sqrt{2}} = \frac{\mu_2}{\mu_1}$. Inverting both sides to find the ratio of the first medium to the second medium yields $\frac{\mu_1}{\mu_2} = \frac{\sqrt{2}}{1}$, which equates to $\sqrt{2}:1$.

The relationship between the refractive index $n$ of a prism, its angle $A$, and the angle of minimum deviation $\delta_{\min}$ is given by the formula:
$n = \frac{\sin \left( \frac{A + \delta_{\min}}{2} \right)}{\sin \left( \frac{A}{2} \right)}$
Given that the refractive index is $\cot \left( \frac{A}{2} \right)$, we can set the expressions equal to each other:
$\frac{\cos \left( \frac{A}{2} \right)}{\sin \left( \frac{A}{2} \right)} = \frac{\sin \left( \frac{A + \delta_{\min}}{2} \right)}{\sin \left( \frac{A}{2} \right)}$
Canceling the common denominator on both sides leads to $\cos \left( \frac{A}{2} \right) = \sin \left( \frac{A + \delta_{\min}}{2} \right)$. Applying the trigonometric identity $\cos \theta = \sin \left( \frac{\pi}{2} - \theta \right)$, the expression becomes $\sin \left( \frac{\pi}{2} - \frac{A}{2} \right) = \sin \left( \frac{A + \delta_{\min}}{2} \right)$. By equating the arguments of the sine functions, we obtain $\frac{\pi}{2} - \frac{A}{2} = \frac{A + \delta_{\min}}{2}$. Multiplying the entire equation by 2 results in $\pi - A = A + \delta_{\min}$, which simplifies to $\delta_{\min} = \pi - 2A$.

A virtual image formed by a convex lens appears on the same side as the object, with a magnification of 3 implying the image distance $v$ is three times the object distance $u$, or $v = 3u$. Because both the object and virtual image are on the same side, the separation distance between them is the difference between their absolute positions, giving $|v - u| = 20$ cm, which simplifies to $2u = 20$ cm, meaning the object is placed at $u = 10$ cm and the image is at $v = 30$ cm. By applying the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ with the appropriate sign convention, where $u = -10$ cm and $v = -30$ cm, we substitute the values to obtain $\frac{1}{-30} - \frac{1}{-10} = \frac{1}{f}$. Solving this equation, $\frac{2}{30} = \frac{1}{f}$ leads to a focal length of $f = 15$ cm.

A convex mirror always forms a virtual, erect, and diminished image for any real object placed in front of it, meaning the magnification $m$ is positive and equal to $1/2$. Given that the radius of curvature is 30 cm, the focal length $f$ is 15 cm. Since the focal length of a convex mirror is always positive, we use $f = 15$ cm in the standard relation $m = \frac{f}{f-u}$. Substituting the known values into this equation, we get
$\frac{1}{2} = \frac{15}{15-u}$
Solving this equation leads to $15 - u = 30$, which results in $u = -15$ cm.

The refraction of light rays at a curved boundary is determined by the relationship between the refractive indices of the media, the radius of the interface, and the respective positions of the object and image. This phenomenon is described by the equation $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$ where $u$ and $v$ represent the object and image distances measured from the surface, while $\mu_1$ and $\mu_2$ are the refractive indices of the incident and refracting media.

For a source in air with $\mu_1 = 1$ located $u = -100$ cm from a convex surface of $\mu_2 = 1.5$ and radius $R = 20$ cm, applying the sign convention requires that the source distance be negative and the radius be positive. Substituting these values into the refraction formula results in $\frac{1.5}{v} - \frac{1}{-100} = \frac{1.5 - 1}{20}$ which simplifies to $\frac{1.5}{v} + 0.01 = 0.025$ Isolating the term with the image position yields $\frac{1.5}{v} = 0.015$, confirming that $v = 100$ cm. This positive value indicates that the image forms 100 cm behind the surface. Since the source is positioned 100 cm in front of the interface and the image is produced 100 cm behind it, the total distance between the object and the image is 200 cm.

The focal length $f$ of a convex lens relates to the image distance $v$ and object distance $u$ through the lens equation $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$. Because errors in experimental measurements are always additive in magnitude, we determine the uncertainty by differentiating this equation to obtain the relationship:
$-\frac{df}{f^2} = -\frac{dv}{v^2} - \frac{du}{u^2}$
Removing the negative signs from every term yields the expression $\frac{df}{f^2} = \frac{dv}{v^2} + \frac{du}{u^2}$. Multiplying both sides by $f^2$ and substituting the given measurement errors $\Delta v$ and $\Delta u$ for the differentials $dv$ and $du$, the resulting expression for the focal length error is $\Delta f = f^2 \left[ \frac{\Delta v}{v^2} + \frac{\Delta u}{u^2} \right]$.

When a light ray strikes a plane mirror, it reflects at an angle equal to its incidence angle, causing the path of the ray to deviate from its original straight-line direction. Given that the angle of reflection is 30 degrees, the law of reflection dictates that the angle of incidence is also 30 degrees.

The total deviation is determined by the following geometric relationship:
$\delta = 180^\circ - 2i$
Substituting the angle of incidence into this expression gives $\delta = 180^\circ - 2(30^\circ)$, which simplifies to $\delta = 120^\circ$. This calculation results in a final deviation of 120 degrees.

The mirror formula, which relates the object distance $u$, image distance $v$, and focal length $f$, is given by the expression $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$. By substituting the given measurements $u = 40$ cm and $v = 120$ cm, the focal length is calculated as $f = 30$ cm. To determine the error in the measurement of the focal length, we differentiate the mirror formula to find the relationship between the errors, which leads to $\frac{df}{f^2} = \frac{dv}{v^2} + \frac{du}{u^2}$.

Since the scale has 20 small divisions per centimeter, the least count, or uncertainty in each distance measurement, is $\Delta u = \Delta v = \frac{1}{20}$ cm. Substituting these values into the error propagation equation, $\Delta f = f^2 \left( \frac{\Delta v}{v^2} + \frac{\Delta u}{u^2} \right)$, yields $\Delta f = 30^2 \left( \frac{1/20}{120^2} + \frac{1/20}{40^2} \right)$. Calculating this expression results in $\Delta f = \frac{1}{32}$ cm. Given that the error is represented as $\frac{1}{K}$ cm, the value of $K$ is 32.

When light propagates through a medium with a refractive index $\mu$, its speed is reduced to $v = c/\mu$. Because the frequency of the light remains constant regardless of the medium, the relationship between speed and wavelength, defined as $v = f\lambda$, dictates that the wavelength must be inversely proportional to the refractive index, or $\lambda \propto 1/\mu$. This variation in wavelength ensures that light waves travel with different effective optical paths when passing through materials of the same thickness but different refractive indices. Since the phase difference accumulated over a distance is determined by the relationship $\Delta Q = \frac{2 \pi}{\lambda_{vac}} \Delta (\mu x)$, the change in refractive index directly modifies the optical path and the resulting phase, confirming that the wavelength's dependence on the medium is indeed the fundamental reason for the shift in phase.