Ray Optics And Optical Instruments Question Answers: NCERT Class 12 Physics

Welcome to the Chapter 9 - Ray Optics And Optical Instruments, Class 12 Physics NCERT Solutions page. Here, we provide detailed question answers for Chapter 9 - Ray Optics And Optical Instruments. The page is designed to help students gain a thorough understanding of the concepts related to natural resources, their classification, and sustainable development.

Our solutions explain each answer in a simple and comprehensive way, making it easier for students to grasp key topics Ray Optics And Optical Instruments and excel in their exams. By going through these Ray Optics And Optical Instruments question answers, you can strengthen your foundation and improve your performance in Class 12 Physics. Whether you’re revising or preparing for tests, this chapter-wise guide will serve as an invaluable resource.

Exercise 1
A:

Size of the candle, h = 2.5 cm

Image size = h’

Object distance, u = −27 cm

Radius of curvature of the concave mirror, R = −36 cm

Focal length of the concave mirror, f = R/2 = -18 cm

Image distance = v

The image distance can be obtained using the mirror formula:

Therefore, the screen should be placed 54 cm away from the mirror to obtain a sharp image.

The magnification of the image is given as:

The height of the candle’s image is 5 cm. The negative sign indicates that the image is inverted and virtual.

If the candle is moved closer to the mirror, then the screen will have to be moved away from the mirror in order to obtain the image.


A:

Focal length of the convex lens, f1 = 30 cm

Focal length of the concave lens, f2 = −20 cm

Focal length of the system of lenses = f

The equivalent focal length of a system of two lenses in contact is given as:

 

1 / f  = 1 / f1 + 1 / f2

1 / f = 1 / 30 - 1 / 20 = 2 - 3 / 60 = - 1 / 60 

So f = - 60 cm

Hence, the focal length of the combination of lenses is 60 cm. The negative sign indicates that the system of lenses acts as a diverging lens.


A:

Height of the needle, h1 = 4.5 cm

Object distance, u = −12 cm

Focal length of the convex mirror, f = 15 cm

Image distance = v

The value of v can be obtained using the mirror formula:

Hence, the image of the needle is 6.7 cm away from the mirror. Also, it is on the other side of the mirror.

The image size is given by the magnification formula:

Hence, magnification of the image, m = h2/h1 = 2.5/4.5 = 0.56

The height of the image is 2.5 cm.

The positive sign indicates that the image is erect, virtual, and diminished.

If the needle is moved farther from the mirror, the image will also move away from the mirror, and the size of the image will reduce gradually.


A:

A myopic or hypermetropic person can also possess the normal ability of accommodation of the eye-lens. Myopia occurs when the eye-balls get elongated from front to back. Hypermetropia occurs when the eye-balls get shortened. When the eye- lens loses its ability of accommodation, the defect is called presbyopia.


A:

The power of the spectacles used by the myopic person, P = −1.0 D

Focal length of the spectacles, f = 1/P = 1/-1x10-2 = -100 cm

Hence, the far point of the person is 100 cm. He might have a normal near point of 25 cm. When he uses the spectacles, the objects placed at infinity produce virtual images at 100 cm. He uses the ability of accommodation of the eye-lens to see the objects placed between 100 cm and 25 cm.

During old age, the person uses reading glasses of power, P' = +2 D

The ability of accommodation is lost in old age. This defect is called presbyopia. As a result, he is unable to see clearly the objects placed at 25 cm.


A:

In the given case, the person is able to see vertical lines more distinctly than horizontal lines. This means that the refracting system (cornea and eye-lens) of the eye is not working in the same way in different planes. This defect is called astigmatism. The person’s eye has enough curvature in the vertical plane. However, the curvature in the horizontal plane is insufficient. Hence, sharp images of the vertical lines are formed on the retina, but horizontal lines appear blurred. This defect can be corrected by using cylindrical lenses.


Frequently Asked Questions about Ray Optics And Optical Instruments - Class 12 Physics

    • 1. How many questions are covered in Ray Optics And Optical Instruments solutions?
    • All questions from Ray Optics And Optical Instruments are covered with detailed step-by-step solutions including exercise questions, additional questions, and examples.
    • 2. Are the solutions for Ray Optics And Optical Instruments helpful for exam preparation?
    • Yes, the solutions provide comprehensive explanations that help students understand concepts clearly and prepare effectively for both board and competitive exams.
    • 3. Can I find solutions to all exercises in Ray Optics And Optical Instruments?
    • Yes, we provide solutions to all exercises, examples, and additional questions from Ray Optics And Optical Instruments with detailed explanations.
    • 4. How do these solutions help in understanding Ray Optics And Optical Instruments concepts?
    • Our solutions break down complex problems into simple steps, provide clear explanations, and include relevant examples to help students grasp the concepts easily.
    • 5. Are there any tips for studying Ray Optics And Optical Instruments effectively?
    • Yes, practice regularly, understand the concepts before memorizing, solve additional problems, and refer to our step-by-step solutions for better understanding.

Exam Preparation Tips for Ray Optics And Optical Instruments

The Ray Optics And Optical Instruments is an important chapter of 12 Physics. This chapter’s important topics like Ray Optics And Optical Instruments are often featured in board exams. Practicing the question answers from this chapter will help you rank high in your board exams.

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