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Light Reflection and Refraction Test - 44

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Light Reflection and Refraction Test - 44
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  • Question 1
    1 / -0
    The image of an object formed by a device is always virtual and diminished. The device may be
    Solution
    Concave lenses always produce images that are upright, virtual, reduced in size, and located on the object's side of the lens.
    Therefore, the image of an object formed by a concave lens is always virtual and diminished.

  • Question 2
    1 / -0
    A convex lens of focal length10 cm is placed at a distance of 12 cm from a wall. How far from the lens should an object be placed so as to form its real image on the wall?
    Solution
    $$f=10cm , v=12cm , u=?$$

    We know, $$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$$
    $$\dfrac{1}{10}=\dfrac{1}{12}-\dfrac{1}{u}$$

    $$u=-60cm$$

    $$\therefore$$The object must be placed at a distance $$60cm$$ in front of the lens.

  • Question 3
    1 / -0
    A diverging lens of 20 cm focal length forms on image 15 cm from the lens. What is the distance of the object from the lens?
    Solution
    Using lens formula as :

    $$\dfrac{1}{v}- \dfrac{1}{u}= \dfrac{1}{f }$$

    $$\dfrac{1}{15}- \dfrac{1}{20}=\dfrac{1}{u}$$

    $$u= 60 \ cm$$
  • Question 4
    1 / -0
    The refractive index of water with respect to air is 1.33. The refractive index of air with respect to water will be
    Solution
    Answer is D.

    Refractive index is the the ratio of the velocity of light in a vacuum to its velocity in a specified medium.
    The refractive index of water with respect to air is 1.33. So, the refractive index of air with respect to water will be the reciprocal.
    Therefore, refractive index of air with respect to water will be 1/1.33 = 0.75.
    Hence, refractive index of air with respect to water will be 0.75.
  • Question 5
    1 / -0
    When rays of light fall on a convex lens, it
    Solution
    When rays of light fall on a convex lens, then the lens refracts(the incident ray undergoes refraction twice before leaving the convex lens) the incident light rays and merges the light rays at one point. The point where all the light rays meet is called its focus. It converges the light rays to a point. So It is also called a converging lens.

  • Question 6
    1 / -0
    If magnification is positive, the nature of the image is:
    Solution
    $$\text{Magnification}=\dfrac{\text{Image  size}}{\text{Object  size}}$$
    According to new Cartesian sign convention, size of height of real and inverted image is considered negative and that of virtual and erect image is considered positive.
    The heigh of the object, being erect is considered positive always.
    So, for positive magnification, the ratio, mentioned above, will be positive. This implies that the image height will also be positive.
    The image will be virtual and erect.
  • Question 7
    1 / -0
    A ray of light passes from air into water. The angle of refraction will be
    Solution
    Answer is B.

    The diagram below shows a light ray undergoing refraction as it passes from air into water. The incident ray is a ray (drawn perpendicular to the wavefronts) that shows the direction that light travels as it approaches the boundary. Similarly, the refracted ray is a ray (drawn perpendicular to the wavefronts) that shows the direction that light travels after it has crossed over the boundary. In the diagram, a normal line is drawn to the surface at the point of incidence. This line is always drawn perpendicular to the boundary. The angle that the incident ray makes with the normal line is referred to as the angle of incidence. Similarly, the angle that the refracted ray makes with the normal line is referred to as the angle of refraction.
    The cause of refraction is a change in light speed; and wherever the light speed changes most, the refraction is greatest. The speed is related to the optical density of a material that is related to the index of refraction of a material. Air is the least dense material (lowest index of refraction value) and water is the denser material (largest index of refraction value). Thus, it would be reasonable that the most refraction occurs for the transmission of light across an air-water boundary.
    Hence, the angle of refraction will be smaller than the angle of incidence.

  • Question 8
    1 / -0
    The twinkling of stars is due to atmospheric
    Solution

    Stars twinkle when we see them from the Earth's surface because we are viewing them through thick layers of turbulent (moving) air in the earth atmosphere.

    Stars appear as tiny dots in the sky; as their light travels through the many layers of the Earth's atmosphere, the light of the star is bent (refracted) many times and in random directions (light is bent when it hits a change in density - like a pocket of cold air or hot air). This random refraction results in the star winking out (it looks as though the star moves a bit, and our eye interprets this as twinkling).

    Option A is correct.

  • Question 9
    1 / -0
    When reflection takes place from a spherical mirror, positions of the object and its image are measured from the ......................
    Solution
    In a spherical mirror, the position of the object or image is always measured from the pole of the mirror.
    The distance between the pole and the object is called the object distance.
    The distance between the pole and the image is called the image distance.

  • Question 10
    1 / -0
    An object is placed $$15\ cm$$ from a diverging mirror of radius of curvature $$20\ cm$$. What is the image magnification produced?
    Solution
    Diverging means convex mirror
    Object distance, $$u = -15\ cm $$ 
    Image distance, $$v$$
    Radius of curvature, $$R = 20\ cm $$ 
    Focal length, $$f = \dfrac{R}{2} = 10\ cm $$ 

    Using mirror formula:
    $$ \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u} $$ 

    $$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{15}$$

    $$\Rightarrow \dfrac{1}{v} = \dfrac{1}{6}$$

    $$ \Rightarrow v = 6\ cm $$ 

    Now, magnification for diverging lens
    $$m = \dfrac{-v}{u} = - \left (\dfrac{6}{-15} \right ) $$ 

    $$ {m = 0.4} $$ 
    Option A is correct answer
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