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Ray Optics and Optical Instruments Test 10
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Ray Optics and Optical Instruments Test 10

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  • Question 1/15
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    When a plane mirror AB is placed horizontally on level ground at a distance of 60 metres from the foot of a tower, the top of the tower and its image in the mirror subtends, an angle of 90° at B. The height of the tower is:

                                             

    Solutions

    The angle subtended at the eye by the top of the tower with respect to the ground is 45 degrees. This happens when the height of the tower is equal to the distance from the eye. The eye is assumed to be very near to the mirror. (Refer below figure).
    Angle subtended by AB = 90 degrees/2 = 45 degrees.
    Therefore, tanθ=h/60​. That is, tan45=h/60​.So, h=60m.
    Hence, the height of the tower is 60 m.

  • Question 2/15
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    A unnumbered wall clock shows time 04 : 25 : 37, where 1st term represents hours, 2nd represents minutes & the last term represents seconds, What time will its image in a plane mirror show.

    Solutions

  • Question 3/15
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    Two plane mirrors of length L are separated by distance L and a man M2 is standing at distance L from the connecting line of mirrors as shown in figure. A man M1 is walking in a straight line at distance 2L parallel to mirrors at speed u, then man M2 at O will be able to see image of M1 for total time :

            

    Solutions


    The light from man M1​ hits the mirror and then reaches the man M2​. The positions of the moving man need to be found when the light from him hits the corners of the mirror to reach the standing man.
    From geometry,
    AO′=L/2​+2(L/2​)
    =3L/2
    ​Similarly, BO′=(3L/2​)+2(3L/2​)
    =9L/2
    Hence AB=(9L/2)-(3L/2)=3/2
    Time taken to cover this distance=3L/u​
    Similarly time taken to cover CD=3L/u​
    Thus total time he is visible is 6L/u​
     

  • Question 4/15
    1 / -0

    A person is standing in a room of width 200 cm. A plane mirror of vertical length 10 cm is fixed on a wall in front of the person. The person looks into the mirror from distance 50 cm. How much width (height) of the wall behind him will he be able to see : (assume that he uses the full mirror)

    Solutions

    Width of the room = 200cm (Given)
    Length of plane mirror = 10cm  (Given)
    Distance of the person looking into mirror = 50cm  (Given)
    In first triangle -
    TanФ = x/200 --- 1
    In second triangle -
    TanФ = 5/50 --- 2
    Equating both the equations -
    5/50 = x/200
    x = 20cm
    Thus, the height of the wall is 2x+10 = 50
    The width of the wall he will be able to see = 50cm

  • Question 5/15
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    Variation of focal length to form a sharp image on retina is called

    Solutions

    The ability of the eye in terms of variation of the focal length to form sharp images on the retina is called accommodation.
     
    In simpler terms, the ability of the eye to change the focus from a near distance or a far distance is called accommodation of the eye.
    It is possible because of the lens which changes its shape. This makes it possible for the eye to keep focus on an object even when its distance from the eye changes.
     

  • Question 6/15
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    Which of the following cannot form real image of a real object ?

    Solutions

    A concave mirror can form both real and virtual images depending upon the distance between object and mirror.

    A convex mirror can only form virtual, erect, and diminished images; therefore, it is true that it can never form a real image of a real object. 


    -Virtual image cannot be formed from a virtual object.
    - When an object is placed between pole and focus, the image formed by the concave mirror is magnified, virtual and erect.
    -When an object is placed beyond the centre of curvature , image of it is formed between centre of curvature and focus which is diminished,real and inverted,so real image of real object can be formed by concave mirror.
    -When the object is virtual, the image formed is real for the concave surface as shown in figure.

  • Question 7/15
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     A transparent sphere of radius R and refractive index m. An object O is placed at a distance x from the pole of the first surface so that a real image is formed at the pole of the exactly opposite surface.

    If x = 2R, then the value of m is

    Solutions

    (μ/v)- (μ/u) = (μ2- μ1)/R
    Or, (μ/2R) + (1/2R) = μ-1/R
    Or, (μ/2) + (1/2) = (μ-1)/1
    Or, (Μ+1)/2= μ-1
    Or, μ+1=2μ-2
     μ=3

     

  • Question 8/15
    1 / -0

     A transparent sphere of radius R and refractive index m. An object O is placed at a distance x from the pole of the first surface so that a real image is formed at the pole of the exactly opposite surface.

    If x = ∞, then the value of m is

    Solutions


    From the symmetry of the fig.ray inside the sphere is parallel to the principal axis. Taking refraction at A
    (μ2/v)−(μ1/u))=(μ2−μ1)/R
    (μ/∞)−(1/−R)=(μ−1)/R
    (1/R)=(μ−1)/R⇒μ−1=1
    μ=2

  • Question 9/15
    1 / -0

     A transparent sphere of radius R and refractive index m. An object O is placed at a distance x from the pole of the first surface so that a real image is formed at the pole of the exactly opposite surface.


    If an object is placed at a distance R from the pole of first surface, then the real image is formed at a distance R from the pole of the second surface. The refractive index m of the sphere is given by

    Solutions


    From the symmetry of the fig.ray inside the sphere is parallel to the principal axis. Taking refraction at A
    (μ2/v)−(μ1/u))=(μ2−μ1)/R
    (μ/∞)−(1/−R)=(μ−1)/R
    (1/R)=(μ−1)/R⇒μ−1=1
    μ=2

  • Question 10/15
    1 / -0

    Column -II shows the optical phenomenon that can be associated with optical components given in column-I. Note that column-I may have more than one matching options in column-II.

  • Question 11/15
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    Statement-I : If a source of light is placed in front of rough wall its image is not seen.

    Statement-II : The wall does not reflect light.

  • Question 12/15
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    Statement-I : As the distance x of a parallel ray from axis increases, focal length decreases

    Statement-II : As x increases, the distance from pole to the point of intersection of reflected ray with principal axis decreases

              

  • Question 13/15
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    Statement-I : When an object dipped in a liquid is viewed normally, the distance between the image and the object is indepedent of the height of the liquid above the object.

    Statement-II : The normal shift is independent of the location of the slab between the object and the observer.

    Solutions

    a) At first the height of the liquid level is obviously dependent because the normal shift will happen according to the depth of the liquid. 
    Drawing a ray diagram keeping a coin at very deep and another close to the surface.We will notice that the earlier one will have a noticeable change of position but the later one will only change a bit. 
    b) The position of the slab is independent because wherever we keep it the incident ray & the emergent ray at last will be parallel (same medium keeping in mind).
     

  • Question 14/15
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    Statement-I : When two plane mirrors are kept perpendicular to each other as shown ( O si the point object), 3 image will be formed.

    Statement-II : In case of multiple reflection, image of one surface can act as an object for the next surface.

                                                             

  • Question 15/15
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    Statement-I : A piece of paper placed at the position of a real image of a virtual object of intense light will burn after sufficient time.

    Statement-II : A virtual object is that point where the incident rays appear to converge and a real image is that point at which reflected/ refracted rays actually converge.

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