 CBSE Sample Papers NCERT Books

# Chapter 9 Ray Optics And Optical Instruments

You will get to learn about the optical instruments in use, spherical mirrors, and their activities in reflecting as well as refracting light, mirror formula, optical fibres, thin lens formula, etc in the ninth chapter of Class 12 Physics. This chapter also teaches you about microscopes and astronomical telescopes and their magnifying powers.

Download pdf of NCERT Solutions for Class Physics Chapter 9 Ray Optics And Optical Instruments

Download pdf of NCERT Examplar with Solutions for Class Physics Chapter 9 Ray Optics And Optical Instruments

### Exercise 1

•  Q1 A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved? Ans: 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. Q2 A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror. Ans: 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. Q10 What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses. Ans: 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. Q25 Does short-sightedness (myopia) or long-sightedness (hypermetropia) imply necessarily that the eye has partially lost its ability of accommodation? If not, what might cause these defects of vision? Ans: 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. Q26 A myopic person has been using spectacles of power −1.0 dioptre for distant vision. During old age he also needs to use separate reading glass of power + 2.0 dioptres. Explain what may have happened. Ans: 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. Q27 A person looking at a person wearing a shirt with a pattern comprising vertical and horizontal lines is able to see the vertical lines more distinctly than the horizontal ones. What is this defect due to? How is such a defect of vision corrected? Ans: 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.

## Recently Viewed Questions of Class 12 Physics

• NCERT Chapter