Light Reflection and Refraction Test

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Light Reflection and Refraction Test - 63

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Question 1
1 / -0

An object 2 cm high is placed at a distance of 16 cm from a concave mirror, which produces a real image 3 cm high. What is the focal length of the mirror ?

A
-9.6 cm

B
-3.6 cm

C
-6.3 cm

D
-8.3 cm

Solution

Here $$h_1$$ = 2 cm, u = -16 cm, $$h_2$$ = -3 cm (Since image is real and inverted.)
$$\because \, m \, = \, \dfrac{h_2}{h_1} \, = \, - \, \dfrac{v}{u} \, \therefore \, v \, = \, \dfrac{-h_2}{h_1} u \, = \,\dfrac{3}{2} \, \times \, (-16) \, = \, -24 \, cm$$
$$\dfrac{1}{f} \, = \, \dfrac{1}{v} \, + \, \dfrac{1}{u} \, =\, -\dfrac{1}{24} \, - \, \dfrac{1}{16} \, = \, \dfrac{-2 \, -3}{48} \, = \, \dfrac{-5}{48};f \, = \, \dfrac{-48}{5} \, = \, -9.6 \, cm$$
Question 2
1 / -0

A rod of length $$10 \ cm$$ lies along the principal axis of a concave mirror of focal length $$10 \ cm$$ in such a way that its end closer to the pole is $$20 \ cm$$ away from the mirror. The length of the image is

A
$$10 \ cm$$

B
$$15 \ cm$$

C
$$2.5 \ cm$$

D
$$5 \ cm$$

Solution

The position of image of nearer end to mirror can be found out using mirror equation,

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

$$\implies \dfrac{1}{v_1}=\dfrac{1}{-20}-\dfrac{1}{-10}$$

$$\implies v_2=-20$$

Similarly position of image of farther end to mirror can be found out as

$$=\dfrac{1}{v_2}=\dfrac{1}{-(10+20)}-\dfrac{1}{-10}$$

$$\implies v_2=-15$$

Hence the length of the image $$=20cm-15cm=5cm$$
Question 3
1 / -0

A convergent beam of light passes through a diverging lens of focal length $$0\cdot2\ m$$ and comes to focus $$0\cdot3\ m$$ behind the lens. The position of the point at which the beam would converge in the absence of the lens is

A
$$0.12\ m$$

B
$$0.6\ m$$

C
$$0.3\ m$$

D
$$0.15\ m$$

Solution

Here, f = -0.2 m, v = +0.3 m
The lens formula
$$\dfrac{1}{v} \, - \, \dfrac{1}{u} \, = \, \dfrac{1}{f}$$
$$\Rightarrow \, \dfrac{1}{u} \, = \, \dfrac{1}{v} \, - \, \dfrac{1}{f} \, = \, \dfrac{1}{0.3} \, + \, \dfrac{1}{0.2} \, = \, \dfrac{0.5}{0.006}$$
$$u \, = \, \dfrac{0.006}{0.5} \, = \, 0.12 \, m$$
Question 4
1 / -0

An object is placed at $$20\ cm$$ form a convex mirror of focal length $$10\ cm.$$ The image formed by a mirror is

A
Real and at $$20cm$$ from the mirror

B
Virtual and at $$20cm$$ from the mirror

C
Virtual and at $$20/3\ cm$$ from the mirror

D
Real and at $$20/3\ cm$$ from the mirror

Solution

$$u=-20\ cm, f=+10\ cm$$ a\so $$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$$
$$\Rightarrow \dfrac{1}{+10}=\dfrac{1}{v}+\dfrac{1}{(-20)}\Rightarrow v=\dfrac{20}{3}\ cm;$$ virtual image.
Question 5
1 / -0

A square wire of side 1 cm is placed perpendicular to the principle axis of a concave mirror of focal length 15 cm at a distance of 20 cm. The area enclosed by the image of the wire is:

A
4 $$cm^2$$

B
6 $$cm^2$$

C
2 $$cm^2$$

D
9 $$cm^2$$
Question 6
1 / -0

An object is placed at a distance of $$5$$ cm from a convex lens of focal length $$10$$ cm, then the image

A
-10

B
-20

C
-30

D
-40

Solution

We know from lens formula

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

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

$$\dfrac{1}{v}=\dfrac{1}{10}-\dfrac{1}{5}$$

$$v=-10 cm$$

As v = -10 the image will be virtual and form in front of the lens.
Question 7
1 / -0

An object is placed at a distance of $$25cm$$ on the axis of a concave mirror, having focal length $$20cm$$. Find the lateral magnification of an image

A
$$-4$$

B
$$-2$$

C
$$4$$

D
$$2$$

Solution

For spherical mirrors,
$$\dfrac{1} {v} + \dfrac{1} {u} = \dfrac{1} {f} $$
Thus putting values of u and f we get v=-100cm

Now, lateral magnification =$$\dfrac {-v} {u} = - 4$$
Question 8
1 / -0

An object is at $$0.06$$ m from a convex lens of focal length $$0.10$$m. Find the position of the image.

A
$$15 m$$

B
$$50 m$$

C
$$15 cm$$

D
$$50 cm$$

Solution

We know from lens formula $$\dfrac { 1 }{ f } =\dfrac { 1 }{ v } -\dfrac { 1 }{ u } $$
where f=focal length=$$0.10m=10$$cm,
u= object distance=$$-0.06m=-6 $$cm and v=image distance
Thus putting the above values in the equation we get,
$$\dfrac { 1 }{ v } =\dfrac { 1 }{ f } -\dfrac { 1 }{ u } \\ =\dfrac { 1 }{ 10 } -\dfrac { 1 }{ 6 } \\ =\dfrac { 3-5 }{ 30 } =\dfrac { -2 }{ 30 } \\ v=\dfrac { -30 }{ 2 } =-15\quad cm$$
Question 9
1 / -0

A candle placed $$25 \,cm$$ form a lens forms an image on a screen placed $$75 \,cm$$ on the other side of the lens. The focal length and type of the lens should be

A
$$+18.75 \,cm$$ and convex lens

B
$$-18.75 \,cm$$ and concave lens

C
$$+20.25 \,cm$$ and convex lens

D
$$-20.25 \,cm$$ and concave lens

Solution

Given,
Object distance, $$u = -25\ cm$$
Image distance, $$v = +75\ cm$$
Focal length, $$f$$

Using lens formula,
$$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$$

$$\dfrac{1}{f} = \dfrac{1}{+75} - \dfrac{1}{-25}$$

$$= \dfrac{1 + 3}{75}$$

$$\dfrac{1}{f} = \dfrac{4}{75}$$

$$f = \dfrac{75}{4} = 18.75\ cm$$

The focal length is positive. So, it is a convex lens.
Question 10
1 / -0

Time taken by the sunlight to pass through a window of thickness 4mm whose refractive index is 1.5 is

A
$$3 \times 10^{-8}s$$

B
$$2 \times 10^{-8}s$$

C
$$3 \times 10^{-11}s$$

D
$$2 \times 10^{-11}s$$

Solution

Given,
Refractive index of medium = 1.5
Thickness of window $$=4 \ mm=4\times 10^{-3} \ m$$
Velocity of light in the window
$$=\dfrac {3\times 10^8}{1.5}ms^{-1}=2\times 10^8 ms^{-1}$$

Hence $$t=\dfrac {4\times 10^{-3}}{2\times 10^8}s=2\times 10^{-11}s$$

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Light Reflection and Refraction Test - 63

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Light Reflection and Refraction Test - 63

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SHARING IS CARING

Question 1

-9.6 cm

-3.6 cm

-6.3 cm

-8.3 cm

Solution

Question 2

$$10 \ cm$$

$$15 \ cm$$

$$2.5 \ cm$$

$$5 \ cm$$

Solution

Question 3

$$0.12\ m$$

$$0.6\ m$$

$$0.3\ m$$

$$0.15\ m$$

Solution

Question 4

Real and at $$20cm$$ from the mirror

Virtual and at $$20cm$$ from the mirror

Virtual and at $$20/3\ cm$$ from the mirror

Real and at $$20/3\ cm$$ from the mirror

Solution

Question 5

4 $$cm^2$$

6 $$cm^2$$

2 $$cm^2$$

9 $$cm^2$$

Question 6

-10

-20

-30

-40

Solution

Question 7

$$-4$$

$$-2$$

$$4$$

$$2$$

Solution

Question 8

$$15 m$$

$$50 m$$

$$15 cm$$

$$50 cm$$

Solution

Question 9

$$+18.75 \,cm$$ and convex lens

$$-18.75 \,cm$$ and concave lens

$$+20.25 \,cm$$ and convex lens

$$-20.25 \,cm$$ and concave lens

Solution

Question 10

$$3 \times 10^{-8}s$$

$$2 \times 10^{-8}s$$

$$3 \times 10^{-11}s$$

$$2 \times 10^{-11}s$$

Solution

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