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

An achromatic prism combination relies on the principle that the net dispersion produced by the system is zero, requiring the dispersive powers of the two prisms to balance such that $\omega_1 \delta_1 = \omega_2 \delta_2$. Since the net deviation of the combination is the difference between the deviations of the individual prisms, we use the relationship $\delta_1 - \delta_2 = 2^\circ$.

Substituting the given dispersive powers $\omega_1 = 0.02$ and $\omega_2 = 0.03$ into the balancing condition, we find that $\delta_2 = \frac{0.02}{0.03} \delta_1 = \frac{2}{3} \delta_1$. Plugging this into the net deviation equation gives $\delta_1 - \frac{2}{3} \delta_1 = 2^\circ$, which simplifies to $\frac{1}{3} \delta_1 = 2^\circ$, resulting in a deviation of $\delta_1 = 6^\circ$ for the crown glass prism. Applying the thin prism deviation formula $\delta_1 = (\mu_1 - 1) A_1$ with the refractive index $\mu_1 = 1.5$, we solve $6^\circ = (1.5 - 1) A_1$ to find that $A_1 = 12^\circ$.

Myopia is a refractive error where the eye is unable to focus properly on distant objects, causing them to appear blurry. When these objects also appear distorted or non-uniform, such as a window mesh looking warped, it points to an additional condition called astigmatism, which is caused by an irregular curvature of the cornea. Since your friend experiences both the lack of clarity at a distance and the specific distortion of the mesh, her vision defect is identified as myopia combined with astigmatism.

The focal length of a thin lens depends on the refractive indices of both the lens material and the surrounding medium according to the Lens Maker's Formula, $\frac{1}{f} = (\frac{\mu_{lens}}{\mu_{medium}} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$. Since the geometric curvature factor $(\frac{1}{R_1} - \frac{1}{R_2})$ remains constant regardless of the medium, the ratio of the focal lengths is determined solely by the inverse of the refractive index terms. In air, the index term is $(\frac{1.5}{1} - 1) = 0.5$, whereas in the liquid with a refractive index of 1.42, the term becomes $(\frac{1.5}{1.42} - 1) \approx 0.056$. Relating these two states shows that $\frac{f_1}{f} = \frac{0.5}{0.056}$, yielding approximately 8.93, which is closest to the integer 9.

A point source of light submerged in a tank emits light rays in all directions, which corresponds to a total solid angle of $4\pi$ steradians. As these rays travel toward the water-air interface, Snell's law dictates that rays incident at an angle greater than the critical angle $\theta_c$ will undergo total internal reflection and remain trapped within the water. For water with a refractive index of $\mu = 4/3$, the critical angle is defined by the relationship $\sin \theta_c = 1/\mu = 3/4$.

The light that manages to escape forms a cone with a half-angle of $\theta_c$, subtending a solid angle of $2\pi(1 - \cos \theta_c)$ steradians. By applying the trigonometric identity $\cos \theta_c = \sqrt{1 - \sin^2 \theta_c}$, we find $\cos \theta_c = \sqrt{1 - (3/4)^2} = \sqrt{7}/4$. The fraction of light that emerges is the ratio of this solid angle to the total sphere, expressed as $\frac{2\pi(1 - \cos \theta_c)}{4\pi} = \frac{1 - \sqrt{7}/4}{2}$. Evaluating this expression yields a value of approximately $0.17$, meaning that nearly $17\%$ of the light escapes into the air.

The behavior of an object in front of a concave mirror is dictated by the fundamental mirror equation $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$, which relates object distance and focal length to image position, combined with the magnification definition $m = -\frac{v}{u}$. When the object is positioned precisely at the focal point where $x = f$, the image is formed at infinity, causing the magnitude of magnification to approach an infinite value. As the object recedes towards the center of curvature at $x = 2f$, the image moves closer to the mirror until both the object and image coincide at that point with a magnification magnitude of $|m| = 1$, describing a continuous curve that drops from infinity to unity.

When a light ray enters a solid glass sphere with a refractive index of $\mu = \sqrt{3}$ at an angle of incidence of $60^{\circ}$, it undergoes refraction according to Snell's law. Applying the law at the point of entry gives $1 \times \sin 60^{\circ} = \sqrt{3} \times \sin \theta$, where $\theta$ is the angle of refraction. Solving this yields $\sin \theta = \frac{1}{2}$, which means the angle of refraction is $\theta = 30^{\circ}$. Because the ray travels through the sphere to the farther surface, the geometry of the sphere ensures the ray strikes that second surface at an angle of incidence equal to the internal refraction angle of $30^{\circ}$.

At this second interface, the ray splits into a reflected component and a refracted component. The reflected ray follows the law of reflection, meaning it exits the surface at an angle of $30^{\circ}$ with respect to the normal. The refracted ray obeys Snell's law again, $\sqrt{3} \times \sin 30^{\circ} = 1 \times \sin \alpha$, where $\alpha$ is the angle of refraction outside the sphere. This calculation results in $\sin \alpha = \frac{\sqrt{3}}{2}$, so the refracted ray makes an angle of $\alpha = 60^{\circ}$ with the normal. Since the reflected and refracted rays lie on opposite sides of the normal, the angle between them is $30^{\circ} + 60^{\circ} = 90^{\circ}$.

The displacement method provides a way to determine the focal length of a convex lens by observing two distinct positions of the lens that yield a sharp real image on a screen placed at a fixed distance from an object. The relationship between the object-screen separation $D$, the distance between the two lens positions $d$, and the focal length $f$ is given by the formula

$f = \frac{D^2 - d^2}{4D}$

Substituting the given values $D = 100$ cm and $d = 40$ cm into this equation yields $f = \frac{100^2 - 40^2}{4 \times 100} = \frac{8400}{400} = 21$ cm. Converting the focal length to meters, we get $f = 0.21$ m, so the optical power is $P = \frac{1}{0.21} = \frac{100}{21}$ D. Comparing this result to the expression $\frac{N}{100}$ D, we set up the equation $\frac{N}{100} = \frac{100}{21}$, which results in $N = \frac{10000}{21} \approx 476.19$. Rounding to the nearest integer as required, the value of $N$ is 476.

The Lens Maker's Formula establishes the relationship between a lens's power, the refractive index of its material, and the radii of curvature of its surfaces. For a double convex lens where both surfaces have a radius $R$, the power is expressed as $P = (\mu-1)(\frac{2}{R})$. A plano-convex lens made of the same material features one flat surface with an infinite radius and one curved surface with radius $R'$, resulting in a power given by $P' = (\mu-1)(\frac{1}{R'})$. Given that the power of this plano-convex lens is $1.5P$, or $\frac{3}{2}P$, we equate $\frac{3}{2}P = (\mu-1)(\frac{1}{R'})$. Substituting the initial expression for $P$ into this equation yields $\frac{3}{2}(\mu-1)(\frac{2}{R}) = (\mu-1)(\frac{1}{R'})$. Dividing both sides by the refractive index term and simplifying the numerical factors results in $\frac{3}{R} = \frac{1}{R'}$, which confirms that the radius of curvature for the plano-convex lens is $\frac{R}{3}$.

The behavior of a plano-convex lens is governed by the Lens-Maker's formula, which connects the refractive index of the glass and the radii of the two surfaces to the focal length:
$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$
In this specific lens, one surface is curved with a radius of 30 cm, and the second surface is perfectly flat. A flat surface has an infinite radius of curvature, making its contribution to the focal length calculation zero because the reciprocal of infinity is effectively 0.

Using a refractive index of 1.5 and the given radius, the formula simplifies to $\frac{1}{f} = (1.5 - 1)\left(\frac{1}{30} - 0\right)$. This arithmetic results in $\frac{1}{f} = 0.5 \times \frac{1}{30}$, or $\frac{1}{f} = \frac{1}{60}$. Solving for the focal length gives a final value of 60 cm.

Refraction of light when traveling from a liquid into air is governed by Snell's Law, which relates the incident and refractive angles to the refractive indices of the media. By identifying the geometric relationship between the jar's radius of 15 cm and the liquid depth of 30 cm, we can determine the angle of refraction $r$ within the liquid. The tangent of this angle is defined by the ratio of the radius to the depth, giving $\tan r = \frac{15}{30} = 0.5$, which geometrically corresponds to a sine value of $\sin r = \frac{1}{\sqrt{5}}$.

Applying Snell's Law at the liquid-air interface, where the light exits into the air at an angle of 45 degrees, the relationship is defined as follows:
$1 \cdot \sin 45^{\circ} = \mu \sin r$
Substituting the known values into this equation yields:
$\frac{1}{\sqrt{2}} = \mu \left( \frac{1}{\sqrt{5}} \right)$
Solving for the refractive index $\mu$ gives $\sqrt{\frac{5}{2}}$, which is approximately 1.581. Given the definition $\frac{N}{100} = \mu$, we have $\frac{N}{100} = 1.581$, resulting in $N = 158.1$, and the required integer value is $N = 158$.

The velocity of an image in a spherical mirror relates to the object's velocity through the square of the magnification, a relationship derived by differentiating the mirror formula with respect to time. When an object moves along the principal axis, the speed of its image is determined by the equation:

$\left| \frac{dv}{dt} \right| = \left( \frac{v}{u} \right)^2 \left| \frac{du}{dt} \right|$

With the object positioned at 30 cm and the image formed at 10 cm, the ratio of image distance to object distance is 10/30, which simplifies to 1/3. Squaring this ratio results in a magnification factor of 1/9. Multiplying this factor by the object's speed of 9 cm/s gives an image speed of $\frac{1}{9} \times 9 = 1$ cm/s.

Minimum strain on the eye occurs when the final image is viewed at infinity, a state known as normal adjustment. In a compound microscope, this condition is met when the intermediate image produced by the objective lens falls exactly at the focal point of the eyepiece. Since the eyepiece has a focal length of 5 cm and the separation between the lenses is 10 cm, the objective lens must form its image 5 cm away.

We apply the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ for the objective lens, where the focal length is 1 cm and the image distance is 5 cm. Substituting these values into the equation yields $\frac{1}{5} - \frac{1}{u} = \frac{1}{1}$. Solving for the object distance gives $-\frac{1}{u} = \frac{4}{5}$, which indicates the magnitude of the distance is 5/4 cm. Since this distance is given as $\frac{n}{40}$ cm, we equate the two expressions to find $n$ using $\frac{n}{40} = \frac{5}{4}$, which results \in $n = 50$.

A compound microscope achieves high magnification by having the objective lens form a real intermediate image, which is then viewed through an eyepiece acting as a simple magnifier. The total magnification is the product of the linear magnification produced by the objective lens and the angular magnification provided by the eyepiece.

Let the object distance for the objective lens be $x$. Applying the lens formula $\frac{1}{v_o} - \frac{1}{-x} = \frac{1}{f_o}$ with a focal length of $f_o = 1$ cm gives an image distance of $v_o = \frac{x}{x-1}$, and the corresponding linear magnification is $m_o = \frac{v_o}{x} = \frac{1}{x-1}$. This intermediate image acts as the object for the eyepiece. Given the tube length of $20$ cm, the distance of this intermediate image from the eyepiece is $|u_e| = 20 - v_o = 20 - \frac{x}{x-1}$.

The eyepiece produces an angular magnification of $m_e = \frac{D}{|u_e|}$, where $D = 25$ cm is the standard near-point distance. Multiplying these values to equate the total magnification to $100$ results in the expression $\left( \frac{1}{x-1} \right) \left( \frac{25}{20 - \frac{x}{x-1}} \right) = 100$. Simplifying the denominator term $20(x-1) - x = 19x - 20$ leads to the equation $\frac{25}{19x-20} = 100(x-1)$, which further reduces to $1 = 4(19x-20)$ or $76x = 81$, yielding $x = \frac{81}{76}$. Substituting this value, we find the object distance for the eyepiece is $|u_e| = \frac{19}{5}$ cm. Finally, using the lens equation for the eyepiece $\frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e}$ with a virtual image distance of $v_e = -25$ cm, we calculate $\frac{1}{f_e} = \frac{1}{-25} - \frac{1}{-19/5} = \frac{106}{475}$, which provides a focal length of approximately $4.48$ cm.

In an astronomical telescope, the tube length is defined by the sum of the focal lengths of the objective and the eyepiece lenses, while the magnifying power represents the ratio between these two focal lengths. Given a tube length of 60 cm and a magnifying power of 5, we apply the relations $f_0 + f_e = 60$ and $\frac{f_0}{f_e} = 5$. By rearranging the second expression as $f_0 = 5f_e$ and substituting this into the first, we obtain $5f_e + f_e = 60$. This simplifies to $6f_e = 60$, indicating that the focal length of the eyepiece is 10 cm.

When light travels from an object at the bottom of a vessel through layers of different liquids, the apparent depth observed from above is the sum of the apparent depths contributed by each individual layer. Each layer shifts the apparent position of the bottom based on the ratio of its physical thickness to its refractive index. Given the vessel consists of two immiscible liquids, each occupying a thickness of h, we calculate the apparent depth of each segment separately.

The lower layer, which has a refractive index of $2\sqrt{2}$, contributes an apparent depth of $\frac{h}{2\sqrt{2}}$. The upper layer, possessing a refractive index of $\sqrt{2}$, contributes an apparent depth of $\frac{h}{\sqrt{2}}$. To find the total apparent depth, we sum these two contributions: $\frac{h}{2\sqrt{2}} + \frac{h}{\sqrt{2}}$. By finding a common denominator, this expression becomes $\frac{h}{2\sqrt{2}} + \frac{2h}{2\sqrt{2}}$, which simplifies to $\frac{3h}{2\sqrt{2}}$. Multiplying the numerator and denominator by $\sqrt{2}$ to rationalize the expression yields a total apparent depth of $\frac{3h\sqrt{2}}{4}$.

The angular resolution of a telescope is defined by Rayleigh's criterion, which sets the minimum angular separation required to distinguish two point sources as $\theta = 1.22 \frac{\lambda}{D}$, where $\lambda$ represents the wavelength of light and $D$ is the aperture diameter. To find the linear separation $x$ on the moon's surface, we multiply this angular resolution by the distance $d$ between the earth and the moon, leading to the relationship $x = \frac{1.22 \times \lambda \times d}{D}$.

To calculate the value, first convert the given measurements into standard SI units: the wavelength $\lambda$ is $5500 \times 10^{-10}$ m and the distance $d$ is $4 \times 10^8$ m. Substituting these values into the equation yields $x = \frac{1.22 \times (5500 \times 10^{-10}) \times (4 \times 10^8)}{5}$. Simplifying the numerical terms gives $x = \frac{1.22 \times 5500 \times 4 \times 10^{-2}}{5}$, which further reduces to $1.22 \times 44$. This results in a calculated minimum separation of 53.68 m, which is closest to the value of 60 m.

The magnification of a compound microscope when the final image is viewed at the least distance of distinct vision is defined by the formula $m = \frac{L}{f_o} \left(1 + \frac{D}{f_e}\right)$ where $m$ is the total magnification, $L$ is the tube length, $f_o$ is the focal length of the objective, $D$ is the near point distance of 250 mm, and $f_e$ is the focal length of the eyepiece. Substituting the given values of $m = 375$, $L = 150 \text{ mm}$, and $f_o = 5 \text{ mm}$ into this equation gives $375 = \frac{150}{5} \left(1 + \frac{250}{f_e}\right)$.

This expression simplifies to $375 = 30 \left(1 + \frac{250}{f_e}\right)$, which further reduces to $12.5 = 1 + \frac{250}{f_e}$. Subtracting 1 from both sides leads to $11.5 = \frac{250}{f_e}$. Solving for the focal length of the eyepiece yields $f_e = \frac{250}{11.5} \approx 21.7 \text{ mm}$. This value is closest to 22 mm.

This problem requires tracking light rays through a two-stage process: first through the convex lens, then by reflection at the plane mirror, and finally back through the lens. Because the object is positioned at $2f$, where the focal length $f = 0.5$ m, the lens initially forms a real image at an equal distance of 1 m on the opposite side.

Since the mirror is placed 2 m behind the lens, this primary image sits 1 m in front of the mirror surface. A plane mirror reflects this, forming a virtual image at an identical distance of 1 m behind the mirror. For the final stage, this virtual image acts as a new object for the lens. Because it is located 3 m from the lens, we assign the object distance $u = -3$ m and the focal length $f = 0.5$ m to account for the light traveling from the mirror back toward the lens.

Applying the lens equation $\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$ leads to $\frac{1}{v} = \frac{1}{0.5} + \frac{1}{-3}$, which simplifies to $\frac{1}{v} = 2 - \frac{1}{3} = \frac{5}{3}$. Solving this yields $v = 0.6$ m. Since the positive value of $v$ indicates the image forms on the side opposite the incoming light from the mirror, the final image is real. Because the lens is 2 m from the mirror, the total distance of this real image from the mirror is 2 m + 0.6 m = 2.6 m.

The image formed by a concave mirror depends on the object's position relative to the center of curvature, which for this mirror is at 8 cm. With the object placed at $u = -10$ cm, which is beyond the center of curvature, basic optics principles dictate that the resulting image will be real, inverted, and unmagnified.

To verify these properties numerically, use the mirror equation $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$, where the focal length is $f = -4$ cm. Solving $\frac{1}{v} = \frac{1}{-4} - \frac{1}{-10}$ leads to $\frac{1}{v} = \frac{-5 + 2}{20} = -\frac{3}{20}$, yielding an image distance of $v = -\frac{20}{3}$ cm. The magnification is then calculated using $m = -\frac{v}{u}$, which results in $m = -\frac{-20/3}{-10} = -\frac{2}{3}$; the negative sign confirms the inverted orientation, and the absolute value being less than one confirms the image is unmagnified.

The refractive index of a medium is defined by the square root of the product of its relative permittivity $\varepsilon_r$ and relative permeability $\mu_r$, establishing a direct link between electromagnetic properties and optical behavior. The critical angle $\theta_C$ is subsequently determined by the relationship $\sin \theta_C = \frac{1}{n}$, where $n$ is the refractive index.

Substituting the given values, the refractive index becomes $n = \sqrt{3 \times \frac{4}{3}}$, which simplifies to $\sqrt{4}$, or 2. Inserting this result into the critical angle formula gives $\sin \theta_C = \frac{1}{2}$, resulting in a critical angle of 30°.