Class 12 Physics Chapter 9 Ray Optics and Optical Instruments

Key Concept: Real-World Application, Critical Angle Analysis

b) Total internal reflection occurs
[Solution Description] To determine what happens, we need to calculate the critical angle using the formula $$\sin(\text{critical angle}) = 1/n$$, where $$n$$ is the refractive index. So, $$\sin(\text{critical angle}) = 1/1.5$$. Calculating the inverse sine gives us approximately 41.8°. When light hits at an angle greater than 41.8°, total internal reflection occurs.

Key Concept: Ray Diagrams

d) The ray passes straight without deviation.
[Solution Description]
When a ray of light passes through the optical center of a convex lens, it continues in a straight line without any deviation because the optical center does not affect its path.

Key Concept: Refractive Index Application

b) 1.42
[Solution Description]
When a lens becomes invisible in a liquid, it means that the refractive indices of the lens and the liquid are equal. Therefore, the refractive index of the lens must be equal to the refractive index of the liquid.

Since the given refractive index of the liquid is 1.42, the refractive index of the lens is also 1.42.

Key Concept: Total Internal Reflection Applications, Refractive Index Calculations

b) Total internal reflection does not occur, angle below critical angle.
[Solution Description] To determine if total internal reflection occurs, we first need to find the critical angle ($$i_c$$) for the interface using the formula:

$$
i_c = \sin^{-1}\left(\frac{n_2}{n_1}\right).
$$

Substituting the given values:

$$
i_c = \sin^{-1}\left(\frac{1.2}{1.5}\right) = \sin^{-1}(0.8).
$$

Solving this gives us approximately:

$$
i_c \approx 53.13^\circ.
$$

Since the angle of incidence (45°) is less than the critical angle (53.13°), total internal reflection does not occur.

Therefore, total internal reflection will NOT happen as the condition for it $$\text{(angle of incidence} > i_c)$$ is not satisfied.

Key Concept: Refractive Index

b) 1.333
[Solution Description] The refractive index $$ n_{12} $$ of medium 2 with respect to medium 1 can be found using the relation:
$$
n_{12} = \frac{1}{n_{21}}
$$
Given $$ n_{21} = 0.75 $$, then:
$$
n_{12} = \frac{1}{0.75} = 1.3333
$$

Key Concept: Tube Length Calculation

b) 130 cm
[Solution Description]
The tube length $$ L $$ of an astronomical telescope is calculated by summing the focal lengths of the objective and the eyepiece:

$$
L = f_o + f_e = 120\,\text{cm} + 10\,\text{cm} = 130\,\text{cm}
$$

Therefore, the tube length is 130 cm.

Key Concept: Basic Concept

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To find the power of the lens, use the formula for power $$P = \frac{1}{f}$$, where $$f$$ is the focal length in meters.

Given:

$$f = 0.25 \text{ meters}$$

Using the formula,

$$P = \frac{1}{0.25}$$

Calculate:

$$P = 4 \text{ dioptres}$$

Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Application of Concepts

b) It will decrease the magnification of the image.

[Solution Description] In a compound microscope, the total magnification $$m$$ is given by $$m = \frac{L}{f_o} \cdot \frac{D}{f_e}$$, where $$f_e$$ is the focal length of the eyepiece. Increasing $$f_e$$ decreases the term $$\frac{D}{f_e}$$, thereby decreasing the total magnification since it would result in less enlargement of the virtual image created by the eyepiece. Consequently, the final image size will be smaller.

 Key Concept: Snell’s Law

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] According to Snell’s Law, $$\frac{\sin i}{\sin r} = n_{21}$$, where $$i$$ is the angle of incidence and $$r$$ is the angle of refraction. Since the refractive index $$n_{21}$$ of glass relative to air is greater than 1, it implies that for a given angle of incidence $$i$$, the angle of refraction $$r$$ in glass will be smaller because $$\sin i > \sin r$$. Hence, both the assertion and reason are true, and the reason correctly explains why the refracted angle is smaller when light enters a denser medium like glass.

Key Concept: Total Internal Reflection Concept

b) denser; rarer
[Solution Description]
Total internal reflection occurs when light travels from a denser medium to a rarer medium, and only if the angle of incidence exceeds the critical angle.

Key Concept: Real vs Virtual Image

b) Virtual
[Solution Description] A magnifying glass primarily forms a virtual image. This occurs because it allows light rays to appear as if they are coming from a larger, more distant object, enabling the eye to focus on them comfortably. Real images, in contrast, can be projected onto a screen, which doesn’t occur with typical use of a magnifying glass aimed for vision magnification.

Key Concept: Real-World Application, Light’s Role in Optical Instruments

c) Its capability of total internal reflection

[Solution Description]

Optical fibers utilize the principle of total internal reflection, which requires that light remains confined within the core material without significant loss. The crucial property here is light’s ability to travel in a straight line combined with the phenomenon of reflection.

Total internal reflection occurs when light traveling in a denser medium hits a less dense interface at an angle greater than the critical angle. Importantly, light must not bend or scatter excessively, which ensures it travels efficiently along the fiber’s path.

Thus, the most critical property of light for maintaining signal integrity is its linear propagation coupled with high-speed transmission.

Key Concept: Optical Fibre Function

b) By reflecting light internally
[Solution Description] Optical fibres use total internal reflection to guide light along their length. The light reflects internally within the fibre with minimal loss, allowing effective transmission over long distances.

Key Concept: Advanced Ray Diagram, Conceptual Complexity

c) Image is real, inverted, enlarged
[Solution Description]
Let’s consider an object placed at a point halfway between the focus ($$F$$) and center of curvature ($$C$$) of a concave mirror.

From ray diagram rules:
– A ray parallel to the principal axis passes through the focal point after reflection.
– A ray passing through the center of curvature reflects back along the same path.
– A ray through the focal point becomes parallel to the principal axis after reflection.

Placing the object midway means it’s positioned at distance $$(f < u < R)$$. Characteristics of the image: - **Position**: Beyond $$C$$ - **Nature**: Real and inverted - **Size**: Larger than the object As per construction, the image forms beyond the center of curvature and is real, inverted, and enlarged.

Key Concept: Comparative Analysis

c) Concave mirror
[Solution Description] Shaving mirrors typically use concave mirrors because they can produce a magnified image when the object is positioned between the focal point and the mirror. This makes it easier for users to see details while shaving.

 Key Concept: Advanced Functioning, Complex Defect Analysis

d) Astigmatism and presbyopia
[Solution Description] The patient’s symptoms suggest issues with focusing on both near and far objects. This can be due to a combination of presbyopia, which affects near vision due to age-related stiffness of the lens, and astigmatism, which causes blurred vision at all distances due to an irregular cornea shape. The other options combine defects that don’t typically cause simultaneous near and distance vision problems.

Key Concept: Angular Magnification

c) 2.5
[Solution Description] The formula for angular magnification when the image is at infinity is $$m = \frac{D}{f}$$, where $$D$$ is the least distance of distinct vision (typically 25 cm), and $$f$$ is the focal length.

Given:
\begin{align*}
f &= 10 \text{ cm} \\
D &= 25 \text{ cm}
\end{align*}

We apply the formula:
$$
m = \frac{25}{10} = 2.5
$$

Therefore, the angular magnification is 2.5.

 Key Concept: Power of Lenses

b) -2D
[Solution Description] The net power (P) of a combination of lenses in contact can be found using the formula:
$$
P = P_1 + P_2
$$
Here, $$ P_1 = +3D $$ and $$ P_2 = -5D $$. Substituting these values:
$$
P = 3 + (-5) = -2D
$$
Thus, the net power is $$-2D$$.

 Key Concept: Sign Convention Understanding

b) Negative
[Solution Description]
In the Cartesian sign convention, distances measured in the direction of the incident light are taken as positive, while those measured against it are negative. If the center of curvature is on the opposite side of the incoming light, $$R$$ is negative.

Key Concept: Objective Lens Challenges

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] The assertion states that large objective lenses in refracting telescopes pose manufacturing and support challenges, which is true because larger lenses require precise glass crafting and robust support structures. The reason provided mentions the heaviness of these lenses and their susceptibility to chromatic aberration. Both statements are accurate since a larger lens indeed has more mass, requiring sturdy support, and chromatic aberration increases with size due to greater light dispersion across lens surfaces. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Optical Fibre Material

b) Core > Cladding
[Solution Description]
For total internal reflection to occur in optical fibers, the refractive index of the core must be greater than that of the cladding. This ensures that the critical angle condition is met at the core-cladding boundary, allowing light to reflect entirely back into the core.

Thus, the relationship should be Core > Cladding.

Key Concept: Real vs Apparent Depth

c) Less than real depth

[Solution Description]

The apparent depth ($$d_a$$) can be calculated using the formula related to real depth ($$d_r$$) and refractive index ($$n$$):

$$d_a = \frac{d_r}{n}$$

Substituting the given values, where $$d_r = 3$$ meters and $$n = 1.33$$, we find:

$$d_a = \frac{3}{1.33} \approx 2.26 \, \text{meters}$$

Therefore, the apparent depth is less than the real depth.

Key Concept: Basic Concept

c) To further magnify the image
[Solution Description] The eyepiece lens in a compound microscope magnifies the image produced by the objective lens, allowing for detailed viewing by the observer.

Key Concept: Basic Focal Length

c) 25 cm
[Solution Description] When an object is placed at infinity, the image is formed at the focal point of the lens. Therefore, the distance between the lens and the image gives us the focal length directly. Thus, the focal length $$f$$ is 25 cm.

Key Concept: Formula Application

c) 55°
[Solution Description] To find the total deviation $$d$$, use the formula $$d = i + e – A$$. Substituting the given values:
$$
d = 40^\circ + 45^\circ – 30^\circ = 55^\circ
$$
So, the total deviation is 55°.

Key Concept: Ray Diagrams

c) At the center of curvature

[Solution Description]

In the case of a concave mirror, when an object is placed at the center of curvature, the rays coming from the object reflect back through the same path since they hit the mirror perpendicular at the point of incidence. Therefore, the image forms at the center of curvature itself. This image is real, inverted, and has the same size as the object.

Key Concept: Application of Sign Convention

b) Negative

[Solution Description] According to the Cartesian sign convention, for mirrors, the distances are measured from the pole of the mirror. The incident light travels towards the mirror, so any real image formed on the same side as the object would have a positive distance. However, in the case of a virtual image formed by a convex mirror, the image appears to be formed behind the mirror. Therefore, the image distance ($$v$$) is taken to be negative.

Key Concept: Angular Magnification

b) $$m_e = \frac{D}{f_e}$$
[Solution Description] When the final image in a compound microscope is formed at infinity, the angular magnification due to the eyepiece is given by the formula:
$$
m_e = \frac{D}{f_e}
$$
Here, $$D = 25 \, \text{cm}$$ is the least distance of distinct vision and $$f_e$$ is the focal length of the eyepiece. Thus, the answer is $$m_e = \frac{D}{f_e}$$.

 Key Concept: Image Formation

c) Behind the primary mirror
[Solution Description] A Cassegrain telescope uses a convex secondary mirror to reflect light back through a hole in the center of the primary mirror. The final image is formed behind the primary mirror after light passes through this opening.

Key Concept: Spectra Formation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] The assertion states that a gas under low pressure produces a line spectrum, which is true. This occurs because the atoms of the gas, when excited, emit light at specific wavelengths, leading to the formation of distinct lines. The reason provided in the statement supports this by explaining the process through which these specific emissions occur. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

 Key Concept: Refractive Index Impact

b) Focal length decreases
[Solution Description] According to the lens maker’s formula:
$$
\frac{1}{f} = \left( \frac{n_2}{n_1} – 1 \right) \left( \frac{1}{R_1} – \frac{1}{R_2} \right)
$$
If the refractive index $$n_2$$ increases and $$n_1$$ remains constant, the quantity $$\left( \frac{n_2}{n_1} – 1 \right)$$ increases. This leads to an increase in $$\frac{1}{f}$$, implying that $$f$$ decreases. Hence, the focal length of the lens will decrease.

Key Concept: Critical Angle Calculation

c) 41.8°
[Solution Description]
To find the critical angle ($$\theta_c$$), we use the formula: $$\sin{\theta_c} = \frac{1}{n}$$, where $$n$$ is the refractive index of glass with respect to air.

Given that the refractive index ($$n$$) of glass is 1.5, we have:
$$
\sin{\theta_c} = \frac{1}{1.5}
$$

Calculating the above expression:
$$
\sin{\theta_c} = 0.6667
$$

Therefore, $$\theta_c = \arcsin(0.6667)$$

Using a calculator, we find:
$$
\theta_c \approx 41.8^\circ
$$
Hence, the critical angle is approximately $$41.8^\circ$$.

Key Concept: Simple Calculation

b) 1.73
[Solution Description]
The refractive index ($$n_{21}$$) is calculated using the formula $$n_{21} = \frac{\sin i}{\sin r}$$, where $$i$$ is the angle of incidence and $$r$$ is the angle of refraction.

$$
n_{21} = \frac{\sin 60^\circ}{\sin 30^\circ} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3} \approx 1.73
$$

Therefore, the refractive index is approximately 1.73.

 Key Concept: Angular Magnification

c) 7.25
[Solution Description]
The formula for angular magnification when the image is at the near point is given by:
$$
m = 1 + \frac{D}{f}
$$
where $$D = 25$$ cm and $$f = 4$$ cm. Substituting these values, we calculate:
$$
m = 1 + \frac{25}{4} = 1 + 6.25 = 7.25
$$
Therefore, the angular magnification is 7.25.

Key Concept: Lens Power

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] A convex lens is known for converging light rays, and its power is measured in dioptres (D). The power of a convex lens is positive because it can focus parallel rays to a point on the retina for farsighted individuals. Hence, both the Assertion and Reason are true, and the reason correctly explains why a convex lens has a positive power.

Key Concept: Lens Design Problem, Real-World Application

b) 15 cm
[Solution Description]
We use the lens maker’s formula:

$$
\frac{1}{f} = (n – 1)\left(\frac{1}{R_1} – \frac{1}{R_2}\right)
$$

Given that the focal length $$f$$ must make the image exactly over the object position because it forms a real image, which implies $$v = u = 60$$ cm, so $$f = 30$$ cm.

With $$R_2 = \infty$$, since it’s plano, the equation boils down to:

$$
\frac{1}{30} = (1.5 – 1)\left(\frac{1}{R_1}\right)
$$

Solving gives:

$$
\frac{1}{30} = 0.5 \cdot \frac{1}{R_1}
$$

Therefore,

$$
R_1 = 15 \text{ cm}
$$

Key Concept: Dispersion and Wavelength, Real-World Application

d) Violet light; because it has a higher refractive index.
[Solution Description]
Dispersion in a prism causes each wavelength to bend by a different amount due to varying refractive indices. Violet light has a higher refractive index (1.53) compared to red light (1.51). According to Snell’s Law, a higher refractive index leads to greater bending or deviation. Hence, violet light will deviate more than red light as it passes through the prism because the change in speed is more significant for shorter (violet) wavelengths due to their higher refractive index.

Key Concept: Advanced Sign Convention, Real World Application

d) -30 cm

[Solution Description] Use the mirror equation:

$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$

Where $$u = -15 \text{ cm}$$ (the object distance is negative as per convention), $$f = -10 \text{ cm}$$.

$$\frac{1}{v} = \frac{1}{f} – \frac{1}{u} = \frac{1}{-10} – \frac{1}{-15}$$

Calculating:

$$\frac{1}{v} = -\frac{3}{30} + \frac{2}{30} = -\frac{1}{30}$$

Thus,

$$v = -30 \text{ cm}$$

The image is virtual and located 30 cm behind the mirror.

 Key Concept: Real vs. Virtual Images

b) Image seen through the eyepiece
[Solution Description]
In a compound microscope, the objective lens forms a real, inverted, and magnified image which is then used by the eyepiece. The eyepiece acts like a simple magnifier, creating a virtual image that is enlarged and can be seen by the observer. Hence, the virtual image is the one formed by the eyepiece.