Light Reflection and Refraction Test

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Light Reflection and Refraction Test - 73

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Weekly Quiz Competition

Question 1
1 / -0

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

A
1 m

B
0.5 m

C
0.24 m

D
2 m

Solution

Here, $$m =\dfrac{v}{u} =-4$$

$$\Rightarrow u = \dfrac{-v}{4}$$------------(1)

Also, $$|u| + |v| = 1.5$$

$$\dfrac{v}{4} + v = 1.5$$

$$\Rightarrow \dfrac{(v+4 v)}{4} = 1.5$$

$$\Rightarrow v = 1.2 m$$

So, putting the value of $$v$$ in equation (1):
$$u= \dfrac{-1.2}{4} = -0.3 m\,\,\,$$

$$\therefore f =\dfrac{uv}{u-v}$$

$$\Rightarrow f=\dfrac{(-0.3 \times 1.2)}{(-0.3 - 1.2)} = 0.24m$$
Hence the correct option is $$(C)$$
Question 2
1 / -0

Refraction of light from air to glass and from air to water are shown in figure (i) and figure (ii) below. The value of the angle $$\theta$$ in the case of refraction as shown in figure (iii) will be

A
$$30^o$$

B
$$35^o$$

C
$$60^o$$

D
$$41^o$$

Solution

Using Snell's law, $$^au_g \, = \, \dfrac{sin \, 60^{\circ}}{sin \, 35^{\circ}}, \, ^a\mu_w \, = \, \dfrac{sin \, 60^{\circ}}{sin \, 41^{\circ}} \, and \, ^w\mu_g \, = \, \dfrac{sin \, 41^{\circ}}{sin \, \theta}$$
$$^a\mu_w \, \times \, ^w\mu_g \, = \, ^q\mu_g$$
$$\Rightarrow \, \dfrac{sin \, 60^{\circ}}{sin \, 41^{\circ}} \, \times \, \dfrac{sin \, 41^{\circ}}{sin \, \theta} \, = \, \dfrac{sin \, 60^{\circ}}{sin \, 35^{\circ}} \, \Rightarrow \, sin \theta \, = \, sin35^{\circ} \, or \, \theta \,= \, 35^{\circ}$$
Question 3
1 / -0

A convex lens of focal length $$f$$ produces a real image $$\mu$$ times the size of an object, Then what is the distance of the object from the lens?

A
$$\left( \mu -1 \right) f$$

B
$$\left( \mu +1 \right) f$$

C
$$\cfrac { \left( \mu -1 \right) f }{ \mu } $$

D
$$\cfrac { \left( \mu +1 \right) f }{ \mu } $$

Solution

Since image formed is real , so magnification will be negative.
Magnification $$m=-\mu$$ $$,-\mu =\dfrac{v}{u}$$ $$,\mu =\dfrac{-v}{u}$$
$$\Rightarrow v=-u\left( \mu\right)$$
Lens formula, $$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$$
$$\Rightarrow \dfrac{-1}{u\left( \mu\right) }-\dfrac{1}{u}=\dfrac{1}{f}$$
$$\Rightarrow u=\dfrac{f\left( \mu +1\right)}{\mu}$$
Hence, the answer is $$\dfrac{f\left( \mu +1\right)}{\mu}.$$
Question 4
1 / -0

The distance between an object and its doubly magnified image by a concave mirror is: [ Assume $$f$$ = focal length]

A
$$ 3 f/2 $$

B
$$2 f/3 $$

C
$$3f$$

D
Depends on whether the image is real or virtual.

Solution

The magnification is given as,
$$m = \dfrac{{ - v}}{u}$$

$$2 = \dfrac{{ - v}}{u}$$

$$v = - 2u$$

Ignoring the sign and using mirror formula, we get

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

$$\dfrac{1}{{2u}} + \dfrac{1}{u} = \dfrac{1}{f}$$

$$\dfrac{{1 + 2}}{{2u}} = \dfrac{1}{f}$$

$$u = \dfrac{{3f}}{2}$$

Here, difference between object distance and image distance is also$$u$$.
Question 5
1 / -0

Refraction of the light from the air to glass and from air to water are shown in the figure (i) and figure (ii) below. The value of an angle $$\theta$$ in the case of refraction as shown in the figure (iii) will be

A
$$30$$

B
$$35$$

C
$$60$$

D
$$41$$

Solution

From snell's law,
From air to glass,
(i) $$\mu_{glass}=\cfrac{\sin 60^{\circ}}{\sin 35^{\circ}}$$

From Air to water,
$$\mu_{water}=\cfrac{\sin 60^{\circ}}{\sin 41^{\circ}}$$

Now, From water to glass
On taking ratio of above equation,
$$\cfrac{\mu_{glass}}{\mu_{water}}=\cfrac{\sin 60^{\circ}}{\sin 35^{\circ}}\cdot \cfrac{\sin 41^{\circ}}{\sin 60^{\circ}}=\cfrac{\sin 41^{\circ}}{\sin 35^{\circ}}$$

Also
$$\cfrac{\mu_{glass}}{\mu_{water}}=\cfrac{\sin 41^{\circ}}{\sin 35^{\circ}}=\cfrac{\sin i}{\sin r}$$

so, where $$i=41^{\circ}, r=35^{\circ}$$
Question 6
1 / -0

One face of a rectangular glass plate $$6{\text{ }}cm$$ thick is silvered. An object held $$8{\text{ }}cm$$ in front of the fist face, forms an image $$12{\text{ }}cm$$ behind the silvered face. The refractive index of the glass is:

A
1.2

B
1.6

C
1.5

D
1.3

Solution

Given: One face of a rectangular glass plate 6 cm thick is silvered. An object held 8 cm in front of the fist face, forms an image 12 cm behind the silvered face.
To find the refractive index of the glass

Solution:

As per the given criteria,

Thickness of glass plate, $$t= 6 cm$$

Distance of the object, $$u = 8 cm$$

And distance of the image, $$v = 12 cm$$

Let $$x$$ = Apparent position of the silvered surface in cm,

Since the image is formed due to reflection at the silvered face and by the property of mirror image

Distance of object from the mirror = Distance of image from the mirror

or, $$x + 8 = 12 + 6 – x \implies x = 5 cm.$$

Therefore, refractive index of glass

$$\mu=\dfrac {\text{Real Depth}}{\text{Apparent Depth}}\\\implies \mu=\dfrac {6}{5}=1.2$$

is the refractive index of the glass.
Question 7
1 / -0

A bulb is located on a wall. Its image is to be obtained on a parallel wall with the help of convex lens. If the distance between parallel walls is $$D$$ then required focal length of lens placed in middle between the walls is :

A
only $$\dfrac { D}{ 4 } $$

B
only $$\dfrac { D }{ 2 } $$

C
More than $$\dfrac { D }{ 4 } $$ but less than $$\dfrac { D }{ 2 } $$

D
less than or equal to $$\dfrac { D }{ 4 } $$

Solution

Refer above image,
$$u=-D/2$$
$$v=D/2$$

$$f=?$$

$$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$$
$$\dfrac{1}{f}=\dfrac{1}{D/2}-\dfrac{1}{-D/2}$$

$$\dfrac{1}{f}=2/D+2/D$$

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

$$f=\dfrac{D}{4}$$
Question 8
1 / -0

An object of length $$6\ cm$$ is placed on the principle axis of a concave mirror of focal length $$f$$ at a distance of $$4\ f$$. The length of the image will be

A
-$$2\ cm$$

B
$$12\ cm$$

C
$$4\ cm$$

D
$$1.2\ cm$$

Solution

Given that,

The object distance $$u=-6\,cm$$

Now, magnification is

$$ m=\dfrac{I}{O} $$

$$ m=\dfrac{f}{f-u} $$

$$ \dfrac{I}{6}=\dfrac{-f}{-f-\left( -4f \right)} $$

$$ I=-2\,cm $$

Hence, the length of image is -$$2\ cm$$
Question 9
1 / -0

The magnification of plane mirror is always -

A
$$ <1 $$

B
$$ > 1 $$

C
$$ = 1 $$

D
Zero

Solution

The size of the image is equal to size of the object for plane mirror.

So, magnification is equal to ratio of image height to object height so it will subsequently be $$1$$ .
Question 10
1 / -0

An infinitely long rod lies along the axis of a concave mirror of focal length l. The near end of the rod is at distance u $$>$$ f from the mirror. Its image will have a length?

A
$$\dfrac{uf}{u-f}$$

B
$$\dfrac{uf}{u+f}$$

C
$$\dfrac{f^1}{u-f}$$

D
$$\dfrac{f^2}{u-f}$$

Solution

The mirror formula is given by the equation
$$\dfrac{1}{f}$$=$$\dfrac{1}{v}$$+$$\dfrac{1}{u}$$---------(1)
where f=focal length, v=image distance and u =object distance
As the condition is for mirror its object distance and focal length both lies on left side of mirror thus we consider it as negative where as its image forms behind the mirror we consider it positive.
Hence equation 1 can be reduced as,
$$\dfrac{1}{v}=\dfrac{1}{u}-\dfrac{1}{f}$$
$$v=\dfrac{uf}{f-u}$$-------(2)
Now length of the image $$L=|v|-|f|$$
$$\dfrac { uf }{ f-u } -f\\ =\dfrac { uf-{ f }^{ 2 }+uf }{ f-u } \\ =\dfrac { { -f }^{ 2 } }{ f-u } $$
$$L=\dfrac { { f }^{ 2 } }{ u-f } $$

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Re-Attempt Weekly Quiz Competition

Light Reflection and Refraction Test - 73

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Light Reflection and Refraction Test - 73

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SHARING IS CARING

Question 1

1 m

0.5 m

0.24 m

2 m

Solution

Question 2

$$30^o$$

$$35^o$$

$$60^o$$

$$41^o$$

Solution

Question 3

$$\left( \mu -1 \right) f$$

$$\left( \mu +1 \right) f$$

$$\cfrac { \left( \mu -1 \right) f }{ \mu } $$

$$\cfrac { \left( \mu +1 \right) f }{ \mu } $$

Solution

Question 4

$$ 3 f/2 $$

$$2 f/3 $$

$$3f$$

Depends on whether the image is real or virtual.

Solution

Question 5

$$30$$

$$35$$

$$60$$

$$41$$

Solution

Question 6

1.2

1.6

1.5

1.3

Solution

Question 7

only $$\dfrac { D}{ 4 } $$

only $$\dfrac { D }{ 2 } $$

More than $$\dfrac { D }{ 4 } $$ but less than $$\dfrac { D }{ 2 } $$

less than or equal to $$\dfrac { D }{ 4 } $$

Solution

Question 8

-$$2\ cm$$

$$12\ cm$$

$$4\ cm$$

$$1.2\ cm$$

Solution

Question 9

$$ <1 $$

$$ > 1 $$

$$ = 1 $$

Zero

Solution

Question 10

$$\dfrac{uf}{u-f}$$

$$\dfrac{uf}{u+f}$$

$$\dfrac{f^1}{u-f}$$

$$\dfrac{f^2}{u-f}$$

Solution

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