Ray Optics and Optical Test

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Ray Optics and Optical Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

A printed page is kept pressed by a glass cube $$(\mu = 1.5)$$ of edge $$6 cm$$. By what amount will the printed letters appear to be shifted when viewed from the top?

A
$$1.5 cm$$

B
$$3 cm$$

C
$$6 cm$$

D
$$2 cm$$

Solution

$$\displaystyle \Delta t=t \left[ 1-\frac {1}{\mu}\right]$$

$$\displaystyle =6\left [ 1-\frac {1}{1.5}\right] = 6 [1-0.67]=6 \times 0.33$$

$$=1.98 \approx 2cm$$

OR

$$\displaystyle RI=\frac {Real depth}{Apparent depth}$$

$$\displaystyle 1.5=\frac {6}{Apparent depth}$$

Apparent depth $$\displaystyle = \frac {6}{1.5}=4$$

So, Shift $$ = 6 - 4 = 2 Cm$$
Question 2
1 / -0

The bottom of a vessel filled with a liquid $$($$refractive index $$= 1.4)$$ appears to be $$15 cm$$ deep when looked vertically downwards. What is the actual depth of the liquid?

A
$$15 cm$$

B
$$30 cm$$

C
$$7.5 cm$$

D
$$21 cm$$

Solution

$$\displaystyle RI=\frac {Real depth}{Apparent depth}$$

$$\displaystyle 1.4=\frac {Real depth}{15 cm}$$

$$\therefore$$ Real depth $$=15 \times 1.4=21cm$$
Question 3
1 / -0

A swimming pool appears to be raised by $$6 m$$ when viewed normally. What is the refractive index of water if the actual depth of the swim-ming pool is $$24 m$$?

A
$$1.44$$

B
$$1.33$$

C
$$1.56$$

D
$$1.76$$

Solution

It is given that surface is rised above by 6m , hence $$apparent depth=24-6=18 m$$ and real depth is given 24 m.
So by formula
$$\mu=\dfrac{real depth}{apparent depth}$$
Hence $$\mu=\dfrac{24}{18}=\dfrac{4}{3}$$
$$\mu=1.33$$
Question 4
1 / -0

The distance between the pole and centre of curvature of a spherical mirror is called _________ .

A
radius of curvature

B
focal length

C
object distance

D
image height

Solution

The radius of the sphere of which the spherical mirror forms a part is known as radius of curvature of the spherical mirror.
Question 5
1 / -0

When the length of a microscope tube increases, its magnifying power

A
decreases

B
increases

C
does not change

D
May increase or decrease depending on the observer and place of observation

Solution

$$\textbf{Explanation:}$$
As we increase the length of microscope tube then we have to increase focal length of the lenses which will decrease its power or magnifying power.
Question 6
1 / -0

A dot on a paper appears to be raised by $$2 cm$$ when viewed normally through a glass slab placed on it. Calculate the thickness of the glass slab if its refractive index is $$\displaystyle \frac {3}{2}$$.

A
$$3 cm$$

B
$$3 m$$

C
$$6 cm$$

D
$$6 m$$

Solution

It is given that vertical shift is 2 cm and we know that
vertical shift $$\delta=t(\mu-1)$$ [OR]
$$\mu=\dfrac{real depth}{apparent depth}$$
and $$apparent depth=R-2$$
Hence $$\dfrac{3}{2}=\dfrac{R}{R-2}$$
$$3R-6=2R$$
$$R=6 cm$$
Hence glass slab is 6 cm thick.
Question 7
1 / -0

LASER belongs to the range of

A
Infrared rays

B
Visible light

C
UV rays

D
X-rays

Solution

nowadays, UV ,Visible and Infrared LASER are also available.
it stands for Light Amplification and Stimulated Emission of Radiation.
Question 8
1 / -0

A point source of light is kept in front of a convex mirror of radius of curvature $$40 cm$$. The image is formed at $$10 cm$$ behind the mirror. Calculate the object distance

A
30

B
20

C
50

D
40

Solution

Given: For a convex mirror, $$R = -40 cm$$.
$$v = -10 cm$$ (image is virtual).
From mirror formula, we have

$$\displaystyle \frac {2}{R}=\frac {1}{u}+\frac {1}{v}$$

$$\displaystyle \frac {1}{u}=\frac {2}{R}-\frac {1}
{v}=\frac {2v-R}{vR}$$

$$\displaystyle u=\frac {vR}{2v-R}=\frac {(-10cm) \times (-40cm)}{2 \times (-10cm) - (-40cm)}$$

$$=+20cm$$
Thus, object is placed $$20 cm$$ in front of the mirror.
Question 9
1 / -0

A student measures the thickness of a human hair by looking at it through a microscope of magnification $$100$$. He makes $$20$$ observations and finds that the average width of the hair in the field of view of the microscope is $$3.5$$ $$mm$$. What is the estimate on the thickness of hair?

A
$$0.035 mm$$

B
$$0.007 mm$$

C
$$3.5 mm$$

D
None of the above

Solution

Magnification= $$100$$.
Avg. width in Microscope =3.5 mm
So,
Original width = $$\cfrac{3.5 \: mm}{100}= 0.035 \: mm$$
Question 10
1 / -0

A thin glass (refractive index 1.5) lens has optical power of $$-5 D$$ in air. Its optical power in a liquid medium with refractive index 1.6 will be:

A
$$-1D$$

B
$$1D$$

C
$$-25D$$

D
$$25D$$

Solution

$$\displaystyle \dfrac{1}{f_a}=\left(\dfrac{1.5}{1}-1\right)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$$........ (i)
$$\displaystyle \dfrac{1}{f_m} = \left(\dfrac{\mu_g}{\mu_m}-1\right)\left(\frac{1}{R_1}-\dfrac{1}{R_2}\right)$$ ....... (ii)
Dividing (i) by (ii), $$\displaystyle \dfrac{f_m}{f_a} = \left(\dfrac{1.5-1}{\dfrac{1.5}{1.6}-1}\right)=-8$$
$$\displaystyle P_a = -5 = \dfrac{\mu}{f_a}=\dfrac{1}{f_a}\Rightarrow f_a = -\dfrac{1}{5}$$
$$\displaystyle \Rightarrow f_m = -8\times f_a = -8\times -\dfrac{1}{5} = \dfrac{8}{5}$$
$$\displaystyle P_m = \dfrac{\mu}{f_m}=\dfrac{1.6}{8} \times 5=1D$$

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Re-Attempt Weekly Quiz Competition

Ray Optics and Optical Test - 34

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Ray Optics and Optical Test - 34

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Question 1

$$1.5 cm$$

$$3 cm$$

$$6 cm$$

$$2 cm$$

Solution

Question 2

$$15 cm$$

$$30 cm$$

$$7.5 cm$$

$$21 cm$$

Solution

Question 3

$$1.44$$

$$1.33$$

$$1.56$$

$$1.76$$

Solution

Question 4

radius of curvature

focal length

object distance

image height

Solution

Question 5

decreases

increases

does not change

May increase or decrease depending on the observer and place of observation

Solution

Question 6

$$3 cm$$

$$3 m$$

$$6 cm$$

$$6 m$$

Solution

Question 7

Infrared rays

Visible light

UV rays

X-rays

Solution

Question 8

30

20

50

40

Solution

Question 9

$$0.035 mm$$

$$0.007 mm$$

$$3.5 mm$$

None of the above

Solution

Question 10

$$-1D$$

$$1D$$

$$-25D$$

$$25D$$

Solution

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