Class 12 Physics NCERT Solutions: Ray Optics and Optical Instruments – Important Questions

Introduction

Class 12 Physics is a crucial stage in understanding fundamental concepts in physics, especially for students preparing for competitive exams and higher education. Among the various topics covered, "Ray Optics and Optical Instruments" is a significant unit that explores the behavior of light and its interaction with different optical devices. This article will provide comprehensive solutions to important questions on ray optics and optical instruments as per the NCERT curriculum. We'll cover key concepts, important questions, and detailed solutions to help students master this topic.

Ray Optics: Key Concepts

Ray optics, also known as geometric optics, deals with the study of light as rays. This approach simplifies the study of light by considering its path as straight lines. The primary concepts in ray optics include reflection, refraction, and the formation of images by lenses and mirrors.

1. Reflection of Light

Reflection occurs when light bounces off a surface. The laws of reflection are:

• First Law of Reflection: The incident ray, the reflected ray, and the normal at the point of incidence all lie in the same plane.
• Second Law of Reflection: The angle of incidence is equal to the angle of reflection.

Important Question:

• Q1: A concave mirror has a focal length of 20 cm. Find the image distance for an object placed 30 cm from the mirror.

Solution: To find the image distance, we use the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ where $f$ is the focal length, $v$ is the image distance, and $u$ is the object distance. Given $f = 20$ cm and $u = -30$ cm (object distance is negative for mirrors), we need to find $v$.

Rearranging the formula: $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ $\frac{1}{v} = \frac{1}{20} - \frac{1}{-30}$ $\frac{1}{v} = \frac{3 - (-2)}{60}$ $\frac{1}{v} = \frac{5}{60}$ $v = \frac{60}{5} = 12 \text{ cm}$

The image distance is 12 cm, and since it is positive, the image is formed on the same side as the object.

2. Refraction of Light

Refraction is the bending of light as it passes from one medium to another. Snell’s Law describes refraction: $n_1 \sin \theta_1 = n_2 \sin \theta_2$ where $n_1$ and $n_2$ are the refractive indices of the two media, and $\theta_1$ and $\theta_2$ are the angles of incidence and refraction, respectively.

Important Question:

• Q2: A light ray travels from air (refractive index = 1) into water (refractive index = 1.33). If the angle of incidence in air is 30°, find the angle of refraction in water.

Solution: Using Snell’s Law: $n_1 \sin \theta_1 = n_2 \sin \theta_2$ $1 \cdot \sin 30^\circ = 1.33 \cdot \sin \theta_2$ $\sin \theta_2 = \frac{\sin 30^\circ}{1.33}$ $\sin \theta_2 = \frac{0.5}{1.33}$ $\sin \theta_2 \approx 0.375$ $\theta_2 \approx \sin^{-1}(0.375) \approx 22.09^\circ$

The angle of refraction in water is approximately 22.09°.

Optical Instruments: Key Concepts

Optical instruments use lenses and mirrors to form images. Understanding how these instruments work is essential for various applications, from microscopes to telescopes.

1. Lenses

Lenses are optical devices made of transparent material that refract light to form images. There are two types of lenses: convex (converging) and concave (diverging).

• Convex Lens: Converges parallel rays to a point called the focus.
• Concave Lens: Diverges parallel rays so that they appear to come from a point called the focus.

The lens formula is: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ where $f$ is the focal length, $v$ is the image distance, and $u$ is the object distance.

Important Question:

• Q3: A convex lens has a focal length of 15 cm. An object is placed 25 cm from the lens. Calculate the image distance and magnification.

Solution: Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ $\frac{1}{15} = \frac{1}{v} - \frac{1}{25}$ $\frac{1}{v} = \frac{1}{15} + \frac{1}{25}$ $\frac{1}{v} = \frac{5 + 3}{75}$ $\frac{1}{v} = \frac{8}{75}$ $v = \frac{75}{8} \approx 9.38 \text{ cm}$

The image distance is approximately 9.38 cm.

Magnification $m$ is given by: $m = \frac{v}{u}$ $m = \frac{9.38}{25} \approx 0.375$

The magnification is approximately 0.375, indicating a diminished image.

2. Microscopes

A microscope is an optical instrument that uses lenses to magnify small objects. The compound microscope consists of two lenses: the objective lens and the eyepiece.

Important Question:

• Q4: A compound microscope has an objective lens of focal length 1 cm and an eyepiece of focal length 5 cm. If the distance between the lenses is 15 cm, calculate the total magnification of the microscope.

Solution: Total magnification $M$ of a compound microscope is given by: $M = \left( \frac{D}{f_o} \right) \left( 1 + \frac{D}{f_e} \right)$ where $f_o$ is the focal length of the objective lens, $f_e$ is the focal length of the eyepiece, and $D$ is the distance between the lenses.

Given $f_o = 1 \text{ cm}$, $f_e = 5 \text{ cm}$, and $D = 15 \text{ cm}$: $M = \left( \frac{15}{1} \right) \left( 1 + \frac{15}{5} \right)$ $M = 15 \left( 1 + 3 \right)$ $M = 15 \cdot 4$ $M = 60$

The total magnification of the microscope is 60x.

3. Telescopes

Telescopes are optical instruments designed to view distant objects by using lenses or mirrors. The primary components of a telescope are the objective lens (or mirror) and the eyepiece.

Important Question:

• Q5: A telescope has an objective lens with a focal length of 100 cm and an eyepiece with a focal length of 2 cm. Calculate the angular magnification of the telescope.

Solution: The angular magnification $M$ of a telescope is given by: $M = \frac{f_o}{f_e}$ where $f_o$ is the focal length of the objective lens and $f_e$ is the focal length of the eyepiece.

Given $f_o = 100 \text{ cm}$ and $f_e = 2 \text{ cm}$: $M = \frac{100}{2}$ $M = 50$

The angular magnification of the telescope is 50x.

Conclusion

Ray optics and optical instruments are fundamental topics in Class 12 Physics. Mastering these concepts and practicing important questions can significantly enhance your understanding and performance in exams. This article has covered key concepts, important questions, and detailed solutions to aid in your preparation. By focusing on reflection, refraction, lenses, microscopes, and telescopes, you can build a strong foundation in ray optics and optical instruments.

For further practice, refer to NCERT textbooks and solve additional problems to ensure a comprehensive grasp of the subject.