Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Geometrical Optics question

2010 · Shift 2 · Q50
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Advanced
  3. /Physics
  4. /Geometrical Optics
  5. /2010 · Shift 2 · Q50

Geometrical Optics question

2010 · Shift 2 · Q50

JEE AdvancedPhysicsGeometrical OpticsNumerical+3 / −1
Image of an object approaching a convex mirror of radius of curvature 20 m along its optical axis is observed to move from 253{{25} \over 3}325​ m to 507{{50} \over 7}750​ m in 30 s. What is the speed of the object in km per hour?
Numerical answer
View written solutionFree

Correct answer: 3

Step-by-step Solution

  1. Identify Given Information and Sign Conventions:

    • The mirror is a convex mirror.
    • Radius of curvature, R = 20 m. For a convex mirror, the radius of curvature is positive, so R = +20 m.
    • Initial image position, v1=253v_1 = {{25} \over 3}v1​=325​ m. For a convex mirror, the image formed is virtual and behind the mirror. According to the Cartesian sign convention, this distance is positive. So, v1=+253v_1 = +{{25} \over 3}v1​=+325​ m.
    • Final image position, v2=507v_2 = {{50} \over 7}v2​=750​ m. Similarly, v2=+507v_2 = +{{50} \over 7}v2​=+750​ m.
    • Time interval, Δt=30\Delta t = 30Δt=30 s.
    • The object is real, so its position u will be negative.
  2. Calculate the Focal Length (f): The focal length of a spherical mirror is half its radius of curvature. f=R2f = {R \over 2}f=2R​ f=+202=+10 mf = {+20 \over 2} = +10 \text{ m}f=2+20​=+10 m

  3. Determine the Initial Object Position (u1): We use the mirror formula: 1v+1u=1f{{1} \over v} + {{1} \over u} = {{1} \over f}v1​+u1​=f1​. For the initial position: 1v1+1u1=1f{{1} \over v_1} + {{1} \over u_1} = {{1} \over f}v1​1​+u1​1​=f1​ 125/3+1u1=110{{1} \over {25/3}} + {{1} \over u_1} = {{1} \over 10}25/31​+u1​1​=101​ 325+1u1=110{{3} \over {25}} + {{1} \over u_1} = {{1} \over 10}253​+u1​1​=101​ 1u1=110−325{{1} \over u_1} = {{1} \over 10} - {{3} \over {25}}u1​1​=101​−253​ To subtract the fractions, we find a common denominator, which is 50. 1u1=550−650=−150{{1} \over u_1} = {{5} \over 50} - {{6} \over 50} = -{{1} \over 50}u1​1​=505​−506​=−501​ u1=−50 mu_1 = -50 \text{ m}u1​=−50 m The negative sign indicates that the object is in front of the mirror, as expected for a real object.

  4. Determine the Final Object Position (u2): Using the mirror formula for the final position: 1v2+1u2=1f{{1} \over v_2} + {{1} \over u_2} = {{1} \over f}v2​1​+u2​1​=f1​ 150/7+1u2=110{{1} \over {50/7}} + {{1} \over u_2} = {{1} \over 10}50/71​+u2​1​=101​ 750+1u2=110{{7} \over {50}} + {{1} \over u_2} = {{1} \over 10}507​+u2​1​=101​ 1u2=110−750{{1} \over u_2} = {{1} \over 10} - {{7} \over {50}}u2​1​=101​−507​ The common denominator is 50. 1u2=550−750=−250=−125{{1} \over u_2} = {{5} \over 50} - {{7} \over 50} = -{{2} \over 50} = -{{1} \over 25}u2​1​=505​−507​=−502​=−251​ u2=−25 mu_2 = -25 \text{ m}u2​=−25 m

  5. Calculate the Distance Traveled by the Object: The object moves from u1=−50u_1 = -50u1​=−50 m to u2=−25u_2 = -25u2​=−25 m. The distance traveled is the magnitude of the displacement. d=∣u2−u1∣=∣−25−(−50)∣=∣−25+50∣=25 md = |u_2 - u_1| = |-25 - (-50)| = |-25 + 50| = 25 \text{ m}d=∣u2​−u1​∣=∣−25−(−50)∣=∣−25+50∣=25 m

  6. Calculate the Speed of the Object: The speed is the distance traveled divided by the time taken. The question asks for

PreviousNext

More from Geometrical Optics

  • Two transparent media of refractive indices μ1​ and μ3​ have a solid lens shaped transparent material of refractive index μ2​ between them as shown in figures in Column II. A ray traversing these media is also shown in the… Includes diagram2010 · MCQ
  • A ball is dropped from a height of 20 m above the surface of water in a lake. The refractive index of water is 4/3. A fish inside the lake, in the line of fall of the ball, is looking at the ball. At an instant, when the ball is 12.8 m…2009 · MCQ
  • A student performed the experiment of determination of focal length of a concave mirror by u-v method using an optical bench of length 1.5 m. The focal length of the mirror used is 24 cm. The maximum error in the location of the image…2009 · Multiple correct
  • Two beams of red and violet colours are made to pass separately through a prism (angle of the prism is 60 ∘). In the position of minimum deviation, the angle of refraction will be :2008 · MCQ
  • A light beam is travelling from Region I to Region IV (Refer figure). The refractive index in Regions I, II, III and IV are n0​,2n0​​,6n0​​ and 8n0​​, respectively. The angle of incidence θ… Includes diagram2008 · MCQ
  • An optical component and an object S placed along its optic axis are given in Column I. The distance between the object and the component can be varied. The properties of images are given in Column II. Match all the properties of images… Includes table Includes diagram2008 · MCQ
  • In an experiment to determine the focal length (f) of a concave mirror by the u-v method, a student places the object pin A on the principal axis at a distance x form the pole P. The student looks at the pin and its inverted image form a…2007 · MCQ
  • A ray of light travelling in water in incident on its surface open to air. The angle of incidence is θ, which is less than the critical angle. Then there will be2007 · MCQ