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

When two thin lenses are separated by a specific distance along a common axis, the reciprocal of the equivalent focal length is calculated using $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}$. Plugging in the provided focal lengths of 30 cm and 10 cm with a separation of 10 cm gives $\frac{1}{F} = \frac{1}{30} + \frac{1}{10} - \frac{10}{10 \times 30}$. Simplifying the terms results in $\frac{1}{F} = \frac{1}{30} + \frac{1}{10} - \frac{1}{30}$, which leaves $\frac{1}{F} = \frac{1}{10}$ per centimeter. Since the power of a lens system is defined as the reciprocal of its focal length in meters, converting this value yields 10 D.

The behavior of light incident on a spherical boundary between two media is described by the refraction formula $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$. Given that the light travels from air with a refractive index $\mu_1 = 1$ into a glass medium with $\mu_2 = 1.5$, and knowing the image distance $v = 200 \text{ cm}$ and the radius of curvature $R = 50 \text{ cm}$, we can substitute these values directly into the governing equation:

$\frac{1.5}{200} - \frac{1}{u} = \frac{1.5 - 1}{50}$

Simplifying the right side of the equation gives $\frac{0.5}{50}$, which is equal to $\frac{1}{100}$. Rearranging the terms to isolate the object distance $u$, we obtain:

$\frac{1}{u} = \frac{1.5}{200} - \frac{1}{100} = \frac{15}{2000} - \frac{20}{2000} = -\frac{5}{2000}$

This calculation results in $\frac{1}{u} = -\frac{1}{400}$, which indicates that the object distance $u$ is $-400 \text{ cm}$. Expressing the magnitude of this distance in meters provides a value of $4 \text{ m}$.

Refraction at a spherical surface is governed by the expression $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2-\mu_1}{R}$, which relates the refractive indices of the two media to the object and image distances. By adopting the standard sign convention where light travels from left to right, we identify the object distance $u$ as -0.2 m and the radius of curvature $R$ as 0.4 m, as the center of curvature lies on the right side.

Substituting these parameters into the refraction formula results in
$\frac{1.5}{v} - \frac{1}{-0.2} = \frac{1.5 - 1}{0.4}$
Simplifying the left side gives $\frac{1.5}{v} + 5$ and the right side simplifies to 1.25. Rearranging these terms leads to the equation
$\frac{1.5}{v} = -3.75$
Solving for $v$ yields -0.4 m. The negative sign confirms that the image forms 0.4 m to the left of the spherical surface.

The power of a lens is inversely proportional to its focal length, which is defined by the Lens Maker's Formula $\frac{1}{f} = (\mu - 1)(\frac{1}{R_1} + \frac{1}{R_2})$. When replacing a lens while keeping its power constant, the sum of the reciprocals of the radii of curvature, expressed as $\frac{1}{R_1} + \frac{1}{R_2}$, must remain invariant. For the initial bi-convex lens with radii of $\frac{1}{6}$ cm for both surfaces, this sum is $6 + 6 = 12$. Applying this requirement to the replacement options, we identify the combination where the reciprocals of the new radii sum to 12. Using radii of $\frac{1}{5}$ cm and $\frac{1}{7}$ cm results in a sum of $5 + 7 = 12$, thereby satisfying the condition to maintain the original lens power.

To determine the focal length of the camera lens, we rely on the principle of magnification, which links the size of the image on the film to the actual size of the object being photographed. Since the camera is at a substantial height, the image distance is approximately equal to the focal length, and the magnification is defined by the ratio of the image side length to the object side length. Given the landscape area is 400 square kilometers, the side length of the object is 20 km, which converts to $20 \times 10^5$ cm. With the film side length given as 2 cm, the magnification is calculated as $m = \frac{2}{20 \times 10^5} = 10^{-6}$. Since the magnification is also equal to the ratio of focal length to object distance, we can use the formula $f = m \cdot u$. Substituting the object distance of $u = 18$ km, which is $18 \times 10^5$ cm, into this relationship yields $f = 10^{-6} \times 18 \times 10^5$ cm, resulting in a focal length of 1.8 cm.

The magnification $m$ produced by a spherical mirror is defined by the ratio of the image distance $v$ to the object distance $u$ as $m = -\frac{v}{u}$. Given a magnification $m = -\frac{1}{3}$, the relationship between the distances is $v = \frac{u}{3}$. The distance between the object and its real image is given by $|u - v| = 30$ cm.

Substituting $v = \frac{u}{3}$ into the distance equation yields $|u - \frac{u}{3}| = 30$, which simplifies to $|\frac{2u}{3}| = 30$. Solving this equation gives $|u| = 45$ cm, and consequently, $|v| = 15$ cm. Since the magnification is negative, the image is real, implying both $u$ and $v$ are negative according to the sign convention, meaning $u = -45$ cm and $v = -15$ cm.

Applying the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ results in $\frac{1}{f} = \frac{1}{-15} + \frac{1}{-45}$. Calculating the right side gives $\frac{1}{f} = \frac{-3 - 1}{45} = -\frac{4}{45}$, which leads to $f = -\frac{45}{4}$ cm. Comparing this to the given expression $f = \frac{x}{4}$ cm, the value of $x$ is $-45$, and the magnitude of $x$ is $45$.

The optical performance of a lens is dictated by the lens maker's formula, which relates the focal length to the refractive indices of the lens and the surrounding medium, as well as the radius of curvature. For a plano-convex lens, this formula is expressed as
$\frac{1}{f} = \left( \frac{n_g}{n_m} - 1 \right) \frac{1}{R}$
where $n_g$ is the refractive index of the glass, $n_m$ is the refractive index of the medium, and $R$ is the radius of curvature of the curved surface.

For the first lens, which is in air with a refractive index of $1$ and a radius of curvature of $2$ cm, the focal length $f_1$ satisfies the equation
$\frac{1}{f_1} = \left( \frac{1.5}{1} - 1 \right) \frac{1}{2} = 0.5 \times \frac{1}{2} = 0.25$
resulting in a focal length of $f_1 = 4$ cm. When the second lens is immersed in a liquid with a refractive index of $1.2$, its focal length $f_2$ is determined using the same principle with a radius of curvature of $3$ cm:
$\frac{1}{f_2} = \left( \frac{1.5}{1.2} - 1 \right) \frac{1}{3} = (1.25 - 1) \frac{1}{3} = 0.25 \times \frac{1}{3} = \frac{1}{12}$
which gives a focal length of $f_2 = 12$ cm. Calculating the ratio of these focal lengths yields $\frac{f_1}{f_2} = \frac{4}{12}$, which simplifies to $1:3$.

When one side of a thin convex lens is polished, the entire system functions effectively as a single concave mirror. Light travels through the lens, reflects off the silvered surface, and passes back through the lens, experiencing refraction twice and reflection once. The equivalent power of this combined system is the sum of the powers of its individual components, expressed as $P_{eq} = 2P_l + P_m$, where $P_l$ is the power of the lens and $P_m$ is the power of the concave mirror formed by the silvered surface. Using the lens maker's formula, the lens power is $P_l = (\mu-1)(\frac{1}{R} - \frac{1}{-R}) = \frac{2(\mu-1)}{R}$, and the mirror power, with radius of curvature $R$, is $P_m = \frac{2}{R}$.

Substituting these expressions leads to the equivalent power calculation:
$P_{eq} = 2\left(\frac{2(\mu-1)}{R}\right) + \frac{2}{R} = \frac{4\mu-4+2}{R} = \frac{2(2\mu-1)}{R}$
The equivalent focal length of the system is the negative reciprocal of the power, $f_{eq} = -\frac{R}{2(2\mu-1)}$. An image forms at the same location as the object when the object is placed at the center of curvature of the equivalent concave mirror, which is twice the magnitude of the equivalent focal length. This results in an object distance of:
$\text{Distance} = \frac{R}{2\mu-1}$

The optical power of a lens is defined as the reciprocal of its focal length, expressed by the relationship $P = \frac{1}{f}$. Since the power increases from 2.5 D to 2.6 D, the corresponding focal lengths are $F_1 = \frac{1}{2.5}$ and $F_2 = \frac{1}{2.6}$. The relative decrease in focal length is calculated by the ratio $\left| \frac{F_2 - F_1}{F_1} \right|$, which simplifies to $\left| \frac{F_2}{F_1} - 1 \right|$. Substituting the inverse relationship between power and focal length into this expression yields $\left| \frac{2.5}{2.6} - 1 \right|$, which equals $\frac{0.1}{2.6}$ and results in a value of approximately 0.04.

When a double convex lens is sliced parallel to its principal axis, as with the $XY$ plane, the curvature and the refractive index of the material remain effectively unchanged for each portion. Consequently, the focal length of the resulting segments, such as $L_3$, remains identical to the focal length ($f$) of the original lens. Because both the original lens and the piece $L_3$ possess the same focal length, their ratio is $1:1$. This differs from a transverse cut along the $AB$ plane, which would increase the focal length of the resulting lens segments to $2f$.

When viewing objects through different transparent media, the refraction at each interface causes the object to appear at a shifted position relative to its actual location. The total apparent shift is the sum of the individual shifts provided by each liquid layer, defined by the expression $h(1 - \frac{1}{n})$ for each layer of thickness $h$ and refractive index $n$.

Applying this principle to the container with two liquid layers, we relate the total shift of 40 cm to the properties of both liquids. For the first layer, the shift is $60(1 - \frac{1}{1.2})$, and for the second layer, the shift is $H(1 - \frac{1}{1.6})$. Summing these, we establish the equation $40 = 60(1 - \frac{1}{1.2}) + H(1 - \frac{1}{1.6})$. Simplifying the fractions, $1 - \frac{1}{1.2}$ evaluates to $\frac{1}{6}$ and $1 - \frac{1}{1.6}$ evaluates to $\frac{3}{8}$. Consequently, the equation becomes $40 = 60(\frac{1}{6}) + H(\frac{3}{8})$, which simplifies to $40 = 10 + \frac{3H}{8}$. Subtracting 10 from both sides leaves $30 = \frac{3H}{8}$, and solving for $H$ results in $H = 80$ cm.

The refractive index $\mu$ of a medium is determined by the speed of light within that material, described by the equation $v = \frac{c}{\mu}$. Since light travels through air at roughly $v \approx c$ but slows down in glass where $v < c$, the refractive index of glass is necessarily higher than that of air, which confirms the assertion. However, optical density is a measure of light propagation properties and does not depend on the mass density of the medium, meaning the reasoning provided is factually incorrect.

Total internal reflection occurs when light traveling from a denser medium to a rarer medium strikes the boundary at an angle greater than or equal to the critical angle, where the critical angle $\theta_c$ is determined by the relationship $\sin \theta_c = \frac{\mu_{\text{surrounding}}}{\mu_{\text{block}}}$. Given that the block has a refractive index of $1.25$, or $\frac{5}{4}$, and the surrounding medium is $1.0$, the sine of the critical angle is $\sin \theta_c = \frac{1}{5/4} = \frac{4}{5}$.

When light enters the flat face of the block at an incident angle $\theta$, it refracts at an angle $\alpha$ relative to the normal inside the block, governed by Snell's law as $1 \times \sin \theta = \frac{5}{4} \sin \alpha$. This implies that $\sin \alpha = \frac{4}{5} \sin \theta$. For total internal reflection to occur at the top surface, the angle of incidence at that interface, which corresponds to $90^\circ - \alpha$, must be at least $\theta_c$. Applying this condition, $\sin(90^\circ - \alpha) \geq \sin \theta_c$ leads to the requirement that $\cos \alpha \geq \frac{4}{5}$. Utilizing the trigonometric identity $\sin^2 \alpha + \cos^2 \alpha = 1$, if $\cos \alpha$ is $\frac{4}{5}$, then $\sin \alpha$ must be $\frac{3}{5}$. Substituting this value back into the refraction equation, we get $\frac{3}{5} = \frac{4}{5} \sin \theta$, which simplifies to $\sin \theta = \frac{3}{4}$, yielding $\theta = \sin^{-1}\left(\frac{3}{4}\right)$.

The equivalent power of a system formed by several thin lenses placed in contact is calculated by taking the sum of the individual powers of each lens. Since the thickness of these lenses is negligible compared to the radii of curvature, we can apply the lens maker's formula to each of the three segments to determine their respective contributions to the total power. Summing these individual components yields the following equation:

$P_{\text{eq}} = \left(\frac{4}{3}-1\right)\left(\frac{1}{R_1}\right) + \frac{1}{2}\left(-\frac{1}{R_1} + \frac{1}{R_2}\right) + \left(\frac{4}{3}-1\right)\left(-\frac{1}{R_2}\right)$

Expanding the terms and simplifying the fractions allows us to regroup the expression based on the reciprocal of the radii. By evaluating the individual parts, the equation becomes:

$P_{\text{eq}} = \frac{1}{3}\left(\frac{1}{R_1}\right) - \frac{1}{2}\left(\frac{1}{R_1}\right) + \frac{1}{2}\left(\frac{1}{R_2}\right) - \frac{1}{3}\left(\frac{1}{R_2}\right)$

Combining the terms associated with $\frac{1}{R_1}$ and $\frac{1}{R_2}$ requires subtracting one-half from one-third, which results in a coefficient of negative one-sixth. Consequently, the combined power of the lens system simplifies to:

$P_{\text{eq}} = -\frac{1}{6}\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

Refraction at a spherical boundary connecting two media is governed by the relationship
$\frac{\mu_B}{v} - \frac{\mu_A}{u} = \frac{\mu_B - \mu_A}{R}$
where light passes from medium A into medium B. Magnification in such systems is defined by the ratio
$m = \frac{v/\mu_B}{u/\mu_A}$

Given the object distance $u = -13\ \mathrm{cm}$ and the magnification $m = -2$ with refractive indices $\mu_A = 1.3$ and $\mu_B = 1.4$, we first determine the image distance $v$ using the magnification formula rearranged as
$v = m \cdot u \cdot \left(\frac{\mu_B}{\mu_A}\right)$
Substituting the known values leads to
$v = -2 \cdot (-13\ \mathrm{cm}) \cdot \left(\frac{1.4}{1.3}\right) = 28\ \mathrm{cm}$

Applying these values to the refraction equation yields
$\frac{1.4}{28} - \frac{1.3}{-13} = \frac{0.1}{R}$
This calculation simplifies to $0.05 + 0.1 = \frac{0.1}{R}$, resulting in $0.15 = \frac{0.1}{R}$. Solving for the radius of curvature gives
$R = \frac{2}{3}\ \mathrm{cm}$

The size of an image produced by a spherical mirror is defined by its magnification, which is a function of the focal length and the object distance. According to the standard sign convention, where light travels toward the mirror from the left, the magnification formula is $m = \frac{f}{f-u}$. Given the radius of curvature is 12 cm, both mirrors have a focal length with a magnitude of 6 cm; specifically, the convex mirror has $f_1 = 6$ cm, while the concave mirror has $f_2 = -6$ cm. With the object placed 18 cm in front of the mirrors, we use the value $u = -18$ cm for both.

Substituting these values into the magnification formula, the convex mirror produces a magnification of $m_1 = \frac{6}{6 - (-18)} = \frac{6}{24} = \frac{1}{4}$. Conversely, the concave mirror produces a magnification of $m_2 = \frac{-6}{-6 - (-18)} = \frac{-6}{12} = -\frac{1}{2}$. To find the ratio of the image sizes, we calculate the absolute ratio of these magnifications, yielding $\left| \frac{m_1}{m_2} \right| = \left| \frac{1/4}{-1/2} \right| = \left| -\frac{1}{2} \right| = \frac{1}{2}$.

The focal length of a lens is determined by the Lens Maker's Formula, which relates the focal length to the refractive indices of the lens material, the surrounding medium, and the curvature of the lens surfaces. When a lens is immersed in a medium with a different refractive index, the effective refractive index ratio changes, which alters the light-bending ability of the lens and shifts its focal length.

For a convex lens in air, the relationship is given by:
$\frac{1}{f} = (\mu - 1) \left( \frac{2}{R} \right)$
Using the given values where the focal length is 24 cm and the refractive index is 1.5, the equation becomes:
$\frac{1}{24} = (1.5 - 1) \left( \frac{2}{R} \right)$
This simplifies to show that the radius of curvature $R$ is 24 cm. When the lens is moved to water with a refractive index of 1.33, the new focal length $f'$ is calculated by:
$\frac{1}{f'} = \left( \frac{1.5}{1.33} - 1 \right) \left( \frac{2}{24} \right)$
Solving this equation leads to a new focal length of 96 cm.

When a light ray enters a parallel-sided glass slab of thickness $h$, it refracts at an angle $r$, traversing a path of length $AB$ within the material equal to $h \sec r$. The lateral shift is the perpendicular distance between the original path of the incident ray and the emergent ray, forming a right-angled triangle where $AB$ is the hypotenuse and the angle opposite the lateral shift is $(i-r)$. By calculating the sine of this angle and multiplying it by the length of the path $AB$, the lateral shift is determined to be $\frac{h \sin(i-r)}{\cos r}$.

The optical behavior of a silvered lens immersed in a liquid is governed by the principle that the system acts as a mirror with a total power equal to twice the refractive power of the lens's curved surface plus the reflective power of the silvered mirror surface. Because the back surface is flat, its reflective power is zero, meaning the system's overall focal power is determined entirely by the refraction through the curved interface.

For a lens of refractive index $\mu_g = 1.5$ immersed in a liquid of refractive index $\mu_l = 1.2$, the refractive power of the curved surface with radius $R$ is given by $\frac{\mu_g - \mu_l}{R}$. Since light travels through this curved surface twice-once upon entering and once after reflecting-the equivalent power of the system is $P = \frac{2(\mu_g - \mu_l)}{R}$. Given that the system acts as a mirror with a focal length $f = 0.2$ m, this power relates to the focal length and the refractive index of the medium via the equation $\frac{\mu_l}{f} = \frac{2(\mu_g - \mu_l)}{R}$.

Substituting the given values into this expression results in $\frac{1.2}{0.2} = \frac{2(1.5 - 1.2)}{R}$. Simplifying the left side gives $6$, and the right side becomes $\frac{0.6}{R}$. Isolating the variable $R$ leads to $R = \frac{0.6}{6}$, which results in a radius of curvature of $0.10$ m.

In a prism, the condition of minimum deviation dictates that light passes symmetrically, making the internal angles of refraction equal to half the prism's vertex angle. Given a prism angle of $60^\circ$, the angle of refraction inside the prism is $30^\circ$. Applying Snell's Law at the first face allows us to relate the angle of incidence $i$ to the refractive index $\mu$ and the angle of refraction $r$:
$\sin i = \mu \sin r$
Substituting $\mu = \sqrt{2}$ and $r = 30^\circ$ yields $\sin i = \sqrt{2} \times \sin(30^\circ)$, which simplifies to $\sin i = \sqrt{2} \times \frac{1}{2} = \frac{1}{\sqrt{2}}$. This evaluation results in an angle of incidence of $45^\circ$.