NCERT Solutions for Class 12 Science Physics Chapter 1 – Ray Optics And Optical Instruments
Dive into the World of Light and Lenses with our Tailored NCERT Solutions for Class 12 Science Physics Chapter 1: Ray Optics And Optical Instruments! Crafted with simplicity and precision, these solutions are the go-to resource for students seeking efficient homework completion and exam readiness. Explore the realm of optics with ease, exclusively on DD Target PMT NCERT Solutions. Page No 345: Question 9.1: 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? ANSWER: 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, 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 real. 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. Page No 345: Question 9.2: 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. ANSWER: 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, 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. Page No 345: Question 9.3: A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again? ANSWER: Actual depth of the needle in water, h1 = 12.5 cm Apparent depth of the needle in water, h2 = 9.4 cm Refractive index of water = μ The value of μcan be obtained as follows: Hence, the refractive index of water is about 1.33. Water is replaced by a liquid of refractive index, The actual depth of the needle remains the same, but its apparent depth changes. Let y be the new apparent depth of the needle. Hence, we can write the relation: Hence, the new apparent depth of the needle is 7.67 cm. It is less than h2. Therefore, to focus the needle again, the microscope should be moved up. ∴Distance by which the microscope should be moved up = 9.4 − 7.67 Page No 345: Question 9.4: Figures 9.34(a) and (b) show refraction of a ray in air incident at 60° with the normal to a glass-air and water-air interface, respectively. Predict the angle of refraction in glass when the angle of incidence in water is 45º with the normal to a water-glass interface [Fig. 9.34(c)]. ANSWER: As per the given figure, for the glass − air interface: Angle of incidence, i = 60° Angle of refraction, r = 35° The relative refractive index of glass with respect to air is given by Snell’s law as: As per the given figure, for the air − water interface: Angle of incidence, i = 60° Angle of refraction, r = 47° The relative refractive index of water with respect to air is given by Snell’s law as: Using (1) and (2), the relative refractive index of glass with respect to water can be obtained as: The following figure shows the situation involving the glass − water interface. Angle of incidence, i = 45° Angle of refraction = r From Snell’s law, r can be calculated as: Hence, the angle of refraction at the water − glass interface is 38.68°. Page No 346: Question 9.5: A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.) ANSWER: Actual depth of the bulb in water, d1 = 80 cm = 0.8 m Refractive index of water, The given situation is shown in the following figure: Where, i = Angle of incidence r = Angle of refraction = 90° Since the bulb is a point source, the emergent light can be considered as a circle of radius, Using Snell’ law, we can write the relation for the refractive index of water as: Using the given figure, we have the relation: ∴R = tan 48.75° × 0.8 = 0.91 m ∴Area of the surface of water = Ï€R2 = Ï€ (0.91)2 = 2.61 m2 Hence, the area of the surface of water through which the light from the bulb can emerge is approximately 2.61 m2. Page No 346: Question 9.6: A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water (refractive …