Light Reflection and Refraction Test

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Light Reflection and Refraction Test - 64

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Question 1
1 / -0

A man with normal near point $$(25 \,cm)$$ reads a book with small point using a magnifying glass, a thin convex lens of focal length $$5 \,cm$$.

A
the closest distance at which he can read the book when viewing through the magnifying glass is $$4.17 \,cm$$

B
the farthest distance at which he can read the books when viewing through the magnifying glass is $$5 \,cm$$

C
the maximum and minimum magnifying power possible using the above simple microscope are 5 & 6.

D
All the above

Solution

This can be solved by using lens formula
$$\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}$$
here f=5cm ,u=-25cm
then v=4.17cm
Question 2
1 / -0

What should be the distance between an object and a concave mirror of focal length of 20 cm so that a virtual image is produced at the distance of 10 cm form the mirror ?

A
12 cm

B
6.6 cm

C
30 cm

D
25 cm

Solution

Given that focal length $$=f=-20cm$$ (concave mirror) and image distance $$v=10cm$$ (virtual image from behind the mirror)
From mirror formula,
$$\dfrac{1}{V}+\dfrac{1}{u}=\dfrac{1}{f}$$
$$\Rightarrow \dfrac{1}{u}=\dfrac{1}{f}-\dfrac{1}{u}=\dfrac{1}{-20}-\dfrac{1}{10}=\dfrac{-1-3}{20}=\dfrac{-3}{20}$$
$$\Rightarrow u=\dfrac{-20}{3}cm.\simeq 6.67cm$$
Question 3
1 / -0

An arrow is placed at a distance of 25 cm from a diverging mirror of focal length 20 cm. Find the image distance.

A
$$12.1 cm$$

B
$$13.1 cm$$

C
$$11.1 cm$$

D
$$14.2 cm$$

Solution

Given,
$$u=-25cm$$ (the minus sign indicates the object is placed in front of the mirror)
$$f=+20cm$$ (plus sign is for a diverging mirror)
$$v=?$$
From the mirror formula,
$$\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}$$

$$\Rightarrow \dfrac{1}{20}=\dfrac{1}{-25}+\dfrac{1}{v}$$

$$\Rightarrow \dfrac{1}{v}=\dfrac{9}{100}$$

$$\Rightarrow v=\dfrac{100}{9}cm$$

$$\therefore v=11.1cm$$

The correct option is C.
Question 4
1 / -0

Light enters at an angle of incidence in a transparent rod of refractive index n. For what value of the refractive index of the material of rod the light once entered into it will not leave it through its lateral face whatsoever be the value of angle of incidence?

A
$$n > \surd 2$$

B
$$n = 1$$

C
$$n = 1.1$$

D
$$n = 1.3$$

Solution

From the following figure
$$r+i=90^o \Rightarrow i=90^o -r$$
For ray not to emerge from curved surface $$i > C$$
$$ \Rightarrow \sin i > \sin C \Rightarrow \sin (90^o -r)> \sin C \Rightarrow \cos r > \sin C$$
$$ \Rightarrow \sqrt {1-\sin^2 r} > \dfrac 1n\quad \left\{\because \sin C=\dfrac 1n \right\}$$
$$ \Rightarrow 1-\dfrac {\sin^2 \alpha}{n^2} > \dfrac {1}{n^2} \Rightarrow 1 > \dfrac {1}{n^2} (1+\sin^2 \alpha)$$
$$ \Rightarrow n^2 > 1 \sin^2 \alpha \Rightarrow n > \sqrt 2\quad \left\{\sin i \to 1\right\}$$
$$ \Rightarrow $$ Least value $$=\sqrt 2$$
Question 5
1 / -0

A fish looking up through the water sees the outyside world contained in a circular horizon. If the refractive index of water is 4/3 and the fish is 12 cm below the surface, the radius of this circle in cm is

A
$$36\sqrt{7}$$

B
$$36/\sqrt{7}$$

C
$$36/\sqrt{5}$$

D
$$4/\sqrt{5}$$

Solution

The refractive index of water is $$N= \dfrac{4}{3}$$
The critical angle is $$i_c = sin^{-1} \dfrac{1}{n}= sin^{-1} \dfrac{3}{4}$$
if $$r$$ is the radius of the circular horizon and $$h$$ is the distance of the fish from the surface then from the geometry, $$cot \ i_c = \dfrac{r}{h}$$

From trigonometry

$$1+ cot^2 i_c = cosec^2 i_c$$

$$1+ \dfrac{h^2}{r^2} = \dfrac{1}{sin ^2 i_c}$$

$$1+ \dfrac{h^2}{r^2} = \dfrac{4^2}{3^2}$$

$$r= \dfrac{h \sqrt 9}{\sqrt 7} = 12 \times \sqrt{ \dfrac{9}{7}} cm = \dfrac{36}{\sqrt 7} cm$$
Question 6
1 / -0

The refractive indices of water and glass are $$\frac{4}{3}\,and\,\frac{3}{2}$$ respectively. The refractive index of water respect to glass is ;

A
$$\frac{8}{9}$$

B
$$2$$

C
$$\frac{2}{3}$$

D
$$\frac{1}{6}$$

Solution

Given,
$$^a\mu_ w=\dfrac{4}{3}$$
$$^a\mu_g=\dfrac{3}{2}$$
The refractive index of water with respect to glass is.
$$^g\mu_w=\dfrac{^a\mu_w}{^a\mu_g}=\dfrac{4/3}{3/2}=\dfrac{8}{9}$$
The correct option is A.
Question 7
1 / -0

Light of frequency $$5\times 10^{14}\ Hz$$ is travelling in a medium of refractive index $$1.5$$. What is its wavelength ? $$(c=3\times 10^{8}\ ms^{-1})$$

A
$$9000\ \mathring { A } $$

B
$$6000\ \mathring { A } $$

C
$$4500\ \mathring { A } $$

D
$$4000\ \mathring { A } $$

Solution
Question 8
1 / -0

A ray is inclident on boundary separating glass and water Refractive index for glass is $$\frac{3}{2}$$ and refractive index for water is $$\frac{4}{3}$$ critical angle for glass-air boundary is

A
$$sin^{-1}$$$$(\frac{3}{4})$$

B
$$sin^{-1}$$$$(\frac{2}{3})$$

C
$$sin^{-1}$$$$(\frac{8}{9})$$

D
$$sin^{-1}$$$$(\frac{1}{9})$$

Solution

$$\textbf{Hint}$$: Apply Snell's law.
$$\textbf{Step 1}$$:
Given a ray is incident on boundary separating glass and water.
Refractive index for glass is $$\dfrac{3}{2}$$ and refractive index for water is $$\dfrac{4}{3}$$ critical angle for glass-air boundary.
For critical angle, the angle of refraction is $$90^{\circ}$$
$$\textbf{Step 2}$$:
Let us use Snell's law. We know from Snell's law $$\mu_1sin{\theta_c}=\mu_2sin90$$
$$\implies \dfrac{3}{2}sin{\theta_c}=\dfrac{4}{3}\times 1$$
$$\implies sin{\theta_c}=\dfrac{8}{9}$$
$$\implies \theta_c=sin^{-1}\dfrac{8}{9}$$
Thus option C is correct
Question 9
1 / -0

A ray of light is incident normally on one of the faces of a prism of apex angle $$30^o$$ and refractive index $$\sqrt 2$$. The angle of deviation of the ray is

A
$$30^o$$

B
$$60^o$$

C
$$15^o$$

D
$$90^o$$

Solution

$$\Delta AQP$$ is $$30 – 60 – 90$$ triangle

Therefore,

$$\theta_1 = 60$$ hence $$\theta_2 = 90 – 60 = 30^0$$

At face $$AC, I = \theta_2 = 30^0$$

Let angle of deviation of the ray be $$r$$ hence snells law at face $$AC$$.

$$\eta = \dfrac{sin\, r}{sin 30} $$

$$sin\, r = n \times sin\,30 = \sqrt 2 \times \dfrac 12 = \dfrac{1}{\sqrt 2} $$

Therefore,

$$r = sin^{–1}\dfrac{1 }{\sqrt 2} = 45^0$$

So,

Angle of deviation $$ = δ = r – 30^0 = 45 – 30 = 15^0$$
Question 10
1 / -0

A convex lens of focal length $$10\ cm$$ forms a vitual image of an object at $$30\ cm$$ from the lens. The magnification produced will be:

A
$$2$$

B
$$3$$

C
$$3.5$$

D
$$4$$

Solution

Given
Focal length, $$f = 10\ cm$$
Image distance $$v = -30\ cm$$
Object distance, $$u$$
Using lens formula,
$$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$$

$$ \implies \dfrac{1}{+10} = \dfrac{1}{\left(-30\right)} - \dfrac{1}{u}$$

$$\implies u = -15\ cm$$

Now, magnification of lens, $$m = \dfrac{v}{u}$$
$$ \implies m = \dfrac{-30}{-15} = 2$$
Hence, $$m = 2$$

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Re-Attempt Weekly Quiz Competition

Light Reflection and Refraction Test - 64

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Light Reflection and Refraction Test - 64

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SHARING IS CARING

Question 1

the closest distance at which he can read the book when viewing through the magnifying glass is $$4.17 \,cm$$

the farthest distance at which he can read the books when viewing through the magnifying glass is $$5 \,cm$$

the maximum and minimum magnifying power possible using the above simple microscope are 5 & 6.

All the above

Solution

Question 2

12 cm

6.6 cm

30 cm

25 cm

Solution

Question 3

$$12.1 cm$$

$$13.1 cm$$

$$11.1 cm$$

$$14.2 cm$$

Solution

Question 4

$$n > \surd 2$$

$$n = 1$$

$$n = 1.1$$

$$n = 1.3$$

Solution

Question 5

$$36\sqrt{7}$$

$$36/\sqrt{7}$$

$$36/\sqrt{5}$$

$$4/\sqrt{5}$$

Solution

Question 6

$$\frac{8}{9}$$

$$2$$

$$\frac{2}{3}$$

$$\frac{1}{6}$$

Solution

Question 7

$$9000\ \mathring { A } $$

$$6000\ \mathring { A } $$

$$4500\ \mathring { A } $$

$$4000\ \mathring { A } $$

Solution

Question 8

$$sin^{-1}$$$$(\frac{3}{4})$$

$$sin^{-1}$$$$(\frac{2}{3})$$

$$sin^{-1}$$$$(\frac{8}{9})$$

$$sin^{-1}$$$$(\frac{1}{9})$$

Solution

Question 9

$$30^o$$

$$60^o$$

$$15^o$$

$$90^o$$

Solution

Question 10

$$2$$

$$3$$

$$3.5$$

$$4$$

Solution

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