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Ray Optics and Optical Instruments Test - 3

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Ray Optics and Optical Instruments Test - 3
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  • Question 1
    4 / -1

    The radii of curvature of the surfaces of a double convex lens are 20cm and 40cm respectively, and its focal length is 20cm. What is the refractive index of the material of the lens?

    Solution

    Here, R1 = 20cm,R2 = −40cm, f = 20cm
    Using lens makers formula we get,


    ⇒ μ = 5/3

  • Question 2
    4 / -1

    A convex lens is dipped in a liquid whose refractive index is equal to the refractive index of the lens. Then its focal length will

    Solution

    When refractive index of lens is equal to the refractive index of liquid, the lens be have lik e a plane surface with focal length infinity.

  • Question 3
    4 / -1

    A convergent beam of light passes through a diverging lens of focal length 0.2m and comes to focus 0.3m behind the lens. The position of the point at which the beam would converge in the absence of the lens is

    Solution

    Here, f = −0.2m,v = +0.3m
    The lens formula



  • Question 4
    4 / -1

    Radii of curvature of a converging lens are in the ratio 1:2. Its focal length is 6cm and refractive index is 1.5. Then its radii of curvature are

    Solution

    Here, f = 6cm, μ = 1.5, R1 = R, R2 = −2R
    According to lens maker's formula

  • Question 5
    4 / -1

    A man is trying to start a fire by focusing sunlight on a piece of paper using an equiconvex lens of focal length 10 cm. The diameter of the sun is 1.39 × 109 m and its mean distance from the earth is 1.5 × 1011m. What is the diameter of the sun's image on the paper?

    Solution

    Magnification for convex lens,

    Here, f = 10cm = 10 × 10−2 m, u = 1.5 × 1011 m

    ∴  Diameter of the image,
    d = m × 1.39 × 109
    = 6.67 × 10−13 × 1.39 × 109
    = 9.27 × 10−4m

  • Question 6
    4 / -1

    A square card of side length 1mm is being seen through a magnifying lens of focal length 10cm. The card is placed at a distance of 9cm from the lens. The apparent area of the card through the lens is

    Solution

    Area of square card = 1 mm × 1 mm = 1 mm2
    Focal length of magnifying lens, f = +10 cm
    Object distance, u = -9 cm
    According to thin lens formula

    or v = -90 cm

    ∴ Apparent area of the card through the  lens 
    = 10 × 10 × 1 mm2
     = 100 mm2
     = 1 cm2

  • Question 7
    4 / -1

    A converging lens is used to form an image on a screen. When the upper half of the lens is covered by an opaque screen

    Solution

    Image formed is complete but has decreased intensity.

  • Question 8
    4 / -1

    Which of the following forms a virtual and erect image for all positions of the object?

    Solution

    Concave lens and convex mirror are diverging in nature. They form virtual and erect images.

  • Question 9
    4 / -1

    A real image of a distance object is formed by a plano-convex lens on its principal axis. Spherical aberration is

    Solution

    When the curved surface of the lens faces the object, the spherical aberration is smaller. The total deviation is shared between the curved and the plane surfaces.

  • Question 10
    4 / -1

    An object is placed at a distance of 1.5 m from a screen and a convex lens is interposed between them. The magnification produced is 4. The focal length of the lens is then

    Solution

    Here, m = u/v = − 4 or u = -v/4
    Also, ∣u∣ + ∣v∣ = 1.5

    or v = 1.2 m and u = -1.2/4 = -0.3 m

  • Question 11
    4 / -1

    The image of the needle placed 45 cm from a lens is formed on a screen placed 90 cm on the other side of the lens. The displacement of the image, if the needle is moved by 5.0 cm away from the lens is 

    Solution

    Here, u = -45 cm,  v = 90 cm 

    when the needle is move 5 cm away from the lens,
    u = -(45 + 5) = -50 cm

    ∴ Displacement of image = v - v' = 90 - 75 = 15 cm, towards the lens

  • Question 12
    4 / -1

    A tree is 18.0 m away and 2.0 m high from a concave lens. How high is the image formed by the given lens of focal length 6m

    Solution

    Focal length of the lens is f = −6.0 m
    u = −18 m and h = 2 m

  • Question 13
    4 / -1

    A luminous object is separated from a screen by distance d. A convex lens is placed between the object and the screen such that it forms a distinct image on the screen. The maximum possible focal length of this convex lens is

    Solution

    From lens displacement method, fmax = d/4

  • Question 14
    4 / -1

    A screen is placed 90cm away from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20cm. Find the focal length of lens.

    Solution

    The image of the object can be located on the screen for two positions of convex lens such that u and v are exchanged.
    The separation between two positions of the lens is d = 20cm

     

    From the figure.
    u1 + v1 = 90 cm
    v1 − u1 = 20 cm
    Solving
    v1 = 55cm, u1 = 35cm
    From lens formula,

  • Question 15
    4 / -1

    A lens having focal length f and aperture of diameter d forms an image of intensity I. Aperture of diameter d/2 in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively

    Solution

    Focal length of the lens remains same.
    Intensity of image formed by lens is proportional to area exposed to incident light from object.
    i.e.  Intensity ∝ area


    After blocking, exposed area,

    Hence, focal length of a lens = f, intensity of the image = 3I/4

  • Question 16
    4 / -1

    A thin convex lens of focal length 25cm is cut into two pieces 0.5cm above the principal axis. The top part is placed at (0, 0) and an object placed at (−50cm, 0). Find the coordinates of the image.

    Solution


    If there was no cut then the object would have been at a height of 0.5cm from the principle axis OO′.
    Consider the image for this case


    Thus the image would have been formed at 50 cm from the pole and 0.5 cm below the principal axis.
    Hence with respect to x-axis passing through the edge of the cut lens, the co-ordinates of the image are (50 cm, −1 cm)

  • Question 17
    4 / -1

    A double convex lens made of glass of refractive index 1.56 has both radii of curvature of magnitude 20cm. If an object is placed at a distance of 10cm from this lens, the position of the image formed is

    Solution

    Here, R1 = 20 cm, R2 = −20 cm, u = −10 cm and μ = 1.56
    Using lens make'rs formula,

    Now, from lens equation,

    Since v is negative, the image will be formed on the same side as that of object.

  • Question 18
    4 / -1

    The power of biconvex lens is 10 dioptre and the radius of curvature of each surface is 10 cm. Then the refractive index of the material of the lens is

    Solution

    Power of lens, P (in dioptre) = 100/f(in cm)
    ∴  f  = 100/10 = 10 cm
    For biconvex lens, R1 = + R, R2 = -R =  According to lens makers formula

  • Question 19
    4 / -1

    A thin glass (refractive index 1.5) lens has optical power of -8 D in air. Its optical power in a liquid medium with refractive index 1.6 will be :

    Solution

    In air,

    In medium

  • Question 20
    4 / -1

    The radius of curvature of each surface of a convex lens of refractive index 1.5 is 40cm. Its power is

    Solution


    Here, μ = 1.5, R1 = 40 cm = 0.4 cm, R2 = −0.4m

    = 2.5D

  • Question 21
    4 / -1

    Two lenses of focal lengths 20cm and −40cm are held in contact. The image of an object at infinity will be formed by the combination at

    Solution

    Equivalent focal length is given by,

    Here, f1 = 20cm and f2 = −40cm

    ⇒ f = 40cm

    ⇒ v = f = 40 cm [∵ u = ∞]

  • Question 22
    4 / -1

    An eye specialist prescribes spectacles having combination of convex lens of focal length 40 cm in contact with a concave lens of focal length 25 cm. The power of this lens combination in diopters is

    (a) +1.5

    Solution

    The focal length of combination is given by

    or 

    ∴ Power of the combination in dioptres, 

    = -1.5

  • Question 23
    4 / -1

    Two lenses of power +10D and −5D are placed in contact. Where should an object be held from the lens, so as to obtain a virtual image of magnification 2?

    Solution

    P1 = +10D, P2 = −5D
    P = P1 + P2 = 10−5 = 5D
    f = 100/5 = 20cm
    Magnification = v/u = 2
    v = 2u




     

  • Question 24
    4 / -1

    A real image of an object is formed at a distance of 20 cm from a lens. On putting another lens in contact with it, the image is shifted 10 cm towards the combination. The power of the lens is

    Solution

    As image formed is real, therefore lens must be convex, v = 20cm.
    Let f1, be the focal length for this lens and f2 be the focal length of the other lens put in contact later.
    for the first lens, 

    After placing it in contact with another lens, the image shifted to 10cm towards the combination
    i.e., v2 = (20 − 10) cm = 10 cm
    Let fnet be the effective focal length of the combination, which can be given as:


    From equation (1) and (2):

    Hence, f2 = 20cm

  • Question 25
    4 / -1

    A concave lens is placed in contact with a convex lens of focal length 25 cm. The combination produces a real image at a distance of 80 cm, If an object is at a distance of 40 cm, the focal length of concave lens is

    Solution

    For the combination of lens,
    Here, u = −40 cm, v = +80 cm,f = ?






    ⇒ f2 = −400cm

  • Question 26
    4 / -1

    Two identical glass (μg = 3/2)  equiconvex lenses of focal length f each are kept in contact. The space between the two lenses is filled with water (μw = 4/3). The focal length of the combination is

    Solution

    Let R be the radius of curvature of each surface. Then

    For the water lens

  • Question 27
    4 / -1

    A convex lens of focal length 15 cm is placed on a plane mirror. An object is placed at 30 cm from the lens. The image is

    Solution


    ⇒ F = 15/2 cm
    ∴ image is real, 10 cm in front of mirror.

  • Question 28
    4 / -1

    In the given figure, the radius of curvature of curved surface for both the plano-convex and plano-concave lens is 10 cm and refractive index for both is 1.5. The location of the final image after all the refractions through lenses is

    Solution

    The focal length of the plano - convex lens is:

    Focal length of plano - concave lens is:

    since parallel beams are incident on the lens, its image from plano - concave lens will be formed at +20 cm from it (at the focus) and will act as an object for the plano - concave lens. since the two lens are at a distance of 10 cm from each other, therefore, for the next lens
    u = +10 cm

  • Question 29
    4 / -1

    A convex lens of radii of curvature 20 cm and 30 cm respectively. It is silvered at the surface which has smaller radius of curvature. Then it will behave as (μg =1.5)

    Solution

    Focal length for lens,

    Here, μg = 1.5, R1 = 20 cm, R2 = −30cm

    Focal length for mirror, 

    ∴ Equivalent focal length

    Hence, this system behaves like a concave mirror of local length -60/11 cm

  • Question 30
    4 / -1

    A concave mirror of focal length f1 is placed at a distance of d from a convex lens of focal length f2. A beam of light coming from infinity and falling on this convex lens - concave mirror combination returns to infinity. The distance d must equal

    Solution


    ∴ d = 2f1 + f2 

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