Light Reflection and Refraction Test

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Light Reflection and Refraction Test - 59

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Question 1
1 / -0

A real object is placed in front of a convex lens of focal length $$f$$ at its principal focus. Then the image is formed at

A
zero

B
infinity

C
at a distance $$2f$$

D
at a distance $$\dfrac f2$$

Solution

The object is placed at focus so $$u=-f$$
For lens formula $$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$$
$$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{-f}$$
$$\dfrac 1v = 0 $$

Hence, $$v=\infty$$, the image is formed at infinity.
Question 2
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A concave mirror gives an image three times as large as the object placed at a distance of 20 cm from it. For the image to be real, the focal length should be.

A
10 cm

B
15 cm

C
20 cm

D
30 cm

Solution

Since the image formed by concave mirror is real ,so it must be an inverted image.
Distance of object from mirror $$u=-20cm$$
Since image is inverted then the magnification will be negative ,hence $$m=-3$$
Magnification $$m=-3=-\dfrac{v}{u}$$
$$m=-3=-\dfrac{v}{-20}$$
$$v=-60cm$$
we know for mirror
$$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$$
$$\dfrac{1}{f}=\dfrac{1}{-60}+\dfrac{1}{-20}$$
$$\dfrac{1}{f}=\dfrac{-60-20}{60\times20}$$=$$\dfrac{-80}{1200}$$=$$\dfrac{-1}{15}$$
$$f=-15cm$$
Question 3
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An object 1 cm tall is placed 4 cm in front of a mirror. In order to produce an upright image of 3 cm height, one needs a

A
convex mirror of radius of curvature 12 cm

B
concave mirror of radius of curvature 12 cm

C
concave mirror of radius of curvature 4 cm

D
plane mirror of height 12 cm

Solution

The concave mirror can produce a magnified upright image.
Distance of object from mirror $$u=-4$$ cm

Magnification $$m=\dfrac{\text{Image height}}{\text{Object height}}=\dfrac31=-\dfrac{v}{u} = -\dfrac{v}{-4}$$

$$v=12cm$$

Now we know for mirror
$$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$$

$$\dfrac{1}{f}=\dfrac{1}{12}+\dfrac{1}{-4}$$

$$f=-6$$ cm

Radius of curvature is $$R=2f=12cm$$ and mirror is concave because the focal length is negative.
Question 4
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Find the size of the image formed by a spherical mirror from the following data :
$$u = -20\ cm$$, $$f = -15\ cm$$ and height of object $$= 1.0\ cm$$

A
$$6\ cm$$

B
$$5\ cm$$

C
$$3\ cm$$

D
$$4\ cm$$

Solution

Given:
$$u=-20$$
$$f=-15$$
We know
$$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$$
$$\dfrac{1}{-15}=\dfrac{1}{v}+\dfrac{1}{-20}$$
$$v=-60cm$$
Now the magnification $$m=-\dfrac{v}{u}=-3$$
Height of object $$h=1cm$$
Height of image $$h_1=|m|\times h=3\times 1=3cm$$
Question 5
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Focal length of a convex lens is $$20 cm$$ and its RI is $$1.5$$. It produces an erect, enlarged image if the distance of the object from the lens is

A
$$40 cm$$

B
$$30 cm$$

C
$$15 cm$$

D
$$20 cm$$

Solution

A convex lens forms an erect and enlarged image when the object is placed between focus and the lens i.e. object distance must be less than the focal length of the lens.
Thus object distance must be $$15 cm$$ (which is less that $$20 cm$$).
Question 6
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A ray of light strikes a glass plate at an angle of $${60}^{o}$$. If the reflected and refracted rays are perpendicular to each other, the refractive index of glass is

A
$$\cfrac{\sqrt{3}}{2}$$

B
$$\cfrac{3}{2}$$

C
$$\cfrac{1}{2}$$

D
$$\sqrt{3}$$

Solution

Let the refractive index of the material be $$\mu$$, and the ange of refraction be $$r$$.
Since the reflected ray and refracted ray are perpendicular to each other, $$60^{\circ}+r=90^{\circ}$$
$$\implies r=30^{\circ}$$
From snell's law,
$$sin i=\mu sinr$$
$$\implies \mu=\dfrac{sin60^{\circ}}{sin30^{\circ}}$$
$$=\sqrt{3}$$
Question 7
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When rays of light are incident on a glass slab, then the incident ray and emergent ray are _________ to each other.

A
Perpendicular

B
Parallel

C
Opposite

D
Concurrent

Solution

The ray of light is incident on the glass slab the ray get refracted twice and comes out from the other surface as the emergent ray.The emergent ray shows a lateral shift from the original path of the light. When the light is passing through the glass its velocity decreases and when comes out of the glass slab its velocity increases and the light ray traces the same direction in the air.
Hence incident and emergent rays are parallel to each other.
Question 8
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Observe the given figure and name the following rays:
Ray AB, Ray BC, Ray CD

A
Incident Ray, Refracted Ray, Emergent Ray

B
Refracted Ray, Incident Ray, Emergent Ray

C
Emergent Ray, Refracted Ray, Incident Ray

D
Refracted Ray, Emergent Ray, Incident Ray

Solution

Ray AB is the incident ray.
Ray BC is the refracted ray
Ray CD is the emergent ray
Question 9
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A glass slab is placed in the path of convergent light. The point of convergence of light :

A
Moves away from slab

B
Moves towards the slab

C
Remains at the same point

D
Undergoes lateral shift

Solution

If the glass slab is placed in the path of convergent light ,the point of convergence or focus will shift away from the glass slab as the light on passing through glass will bend towards the normal and after coming out in air, light will converge out at a point away from the glass slab.
Question 10
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The distance between principal focus and optical center of the lens is ________.

A
Diameter

B
Focal length

C
Principal axis

D
Optical center

Solution

As shown in the figure, the focal length is the distance between the principal focus and the optical center of the lens.