Ray Optics and Optical Test

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Ray Optics and Optical Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

The real depth of a swimming pool that appears to be $10\ m$ for a swimmer if the index of water is $1.33$ is

A
$15.5\ m$

B
$14.3\ m$

C
$13.5\ m$

D
$13.3\ m$

Solution
Question 2
1 / -0

In order to increase the angular magnification of a simple microscope, one should increase

A
The object size

B
The aperture of the lens

C
The focal length of the lens

D
The power of the lens

Solution

When the image is formed at infinity
m=Df=DP
when the image is formed at the near point
m=(1+Df)=1+DP
Question 3
1 / -0

A beam of light is converging towards a point $L$ . A plane parallel plate of glass of thickness $t$ , refractive index $\mu$ is introduced in the path of the beam. The convergent point is shifted by :
(assume near normal incidence).

A
$t\left (1 - \dfrac {1}{\mu}\right ) away$

B
$t\left (1 + \dfrac {1}{\mu}\right ) away$

C
$t\left (1 - \dfrac {1}{\mu}\right ) nearer$

D
$t\left (1 + \dfrac {1}{\mu}\right ) nearer$

Solution

When glass plate of thickness t is introduced, the optical path increases by $\dfrac tn$ .
So the convergence point shifts by $t – \dfrac t\mu$ i.e. $t\left[1 – \dfrac 1\mu\right]$ nearer.
Question 4
1 / -0

The radius curvature for a convex lens is $40\ cm$ , for each surface. Its refractive index is $1.5$ . The local length will be

A
$40\ cm$

B
$20\ cm$

C
$80\ cm$

D
$30\ cm$

Solution

By formula $\dfrac{1}{f}=(\mu-1) \left(\dfrac{1}{R_1}- \dfrac{1}{R_2} \right)$
$=(1.5-1) \left(\dfrac{1}{40}+ \dfrac{1}{40} \right) =0.5 \times \dfrac{1}{20} =\dfrac{1}{40}$
$\therefore f=40\ cm$
Question 5
1 / -0

The time taken by sunlight to cross a $5\ mm$ thick glass plate $(\mu =3/2)$ is

A
$0.25\times 10^{-10}S$

B
$1.67\times 10^{-7}S$

C
$2.5\times 10^{-10}S$

D
$1.167\times 10^{-10}S$

Solution

As we know,
$t=\dfrac {\mu x}{c}=\dfrac {\dfrac {3}{2}\times 5\times 10^{-3}}{3\times 10^8}=0.25\times 10^{-10}s$
Question 6
1 / -0

Finger prints are observed by the use of

A
Telescope

B
Microscope

C
Gallilean telescope

D
Concave lens

Solution

Microscope is used for enlarging tiny objects or details which our eyes cannot detect. It is used to highly magnify the patterns and design of Finger prints and is widely used for Fingerprint tracing.
Question 7
1 / -0

A person using a lens as a simple microscope sees an

A
Inverted virtual image

B
Inverted real magnified image

C
Upright virtual image

D
Upright real magnified image

Solution

A simple microscope is just a convex lens with object lying between optical centre and focus of the lens.
Question 8
1 / -0

A ray of light is incident on the hypotenuse of a right-angled prism after travelling parallel to the base inside the prism. If $\mu$ is the refractive index of the material of the prism, the maximum value of the base angle for which light is totally reflected from the hypotenuse is :

A
$sin^{-1}\left ( \dfrac{1}{\mu } \right )$

B
$tan^{-1}\left ( \dfrac{1}{\mu } \right )$

C
$sin^{-1}\left ( \dfrac{\mu -1}{\mu } \right )$

D
$cos^{-1}\left ( \dfrac{1}{\mu } \right )$

Solution

If $\alpha$ is maximum value of base angle for which light is totally reflected from the hypotenuse and $\theta$ is the minimum value of angle of incidence at hypotenuse for total internal reflection.
Then ,
$\theta =sin^{-1}\left ( \dfrac{1}{\mu}\right )$

$\alpha +\theta =90^o$

$\Rightarrow \alpha = 90^o-sin^{-1}\left ( 1/\mu \right )=cos^{-1}\left ( 1/\mu \right )$
Question 9
1 / -0

The refractive index of the material of a concave lens is $\mu$ and is placed in the medium of refractive index $\mu ^{1}$ . If $\mu ^{1}>\mu$ , then which of the following ray diagram is correct:

A

B

C

D
None of these

Solution

Since, the tendency of a concave lens is to diverge a parallel beam of light falling on the lens surface, it would show exactly the opposite behaviour if placed in a medium of higher refractive index than the material of the lens.
Thus, the correct option will be one in which a concave lens diverges a parallel beam of a light ray.
Question 10
1 / -0

A beam of parallel rays is incident on a transparent slab of refractive index $\sqrt{3}$ making an angle $30^{\circ}$ with the surface of the slab. If the width of incident beam of light is 1.732 mm, the width of refracted beam is:

A
1.00 mm

B
1.50mm

C
2.50 mm

D
3.00 mm

Solution

Using $\mu _{1}sini=\mu _{2}sin\ r$
$1\times sin60^{\circ}=\sqrt{3}\ sin\ r$

$sin\ r=\dfrac{1}{2}$

$AB=\dfrac{1.732mm}{cosi}$

Now,
$BB^{1}=AB\ cos\ r$

$=\dfrac{1.732}{cosi}\times cos\ r$

$=\dfrac{1.732}{\dfrac{1}{2}}\ \times \dfrac{\sqrt{3}}{2}=3\ mm$

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Re-Attempt Weekly Quiz Competition

Ray Optics and Optical Test - 21

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Ray Optics and Optical Test - 21

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Question 1

15.5 m15.5\ m15.5 m

14.3 m14.3\ m14.3 m

13.5 m13.5\ m13.5 m

13.3 m13.3\ m13.3 m

Solution

Question 2

The object size

The aperture of the lens

The focal length of the lens

The power of the lens

Solution

Question 3

t(1−1μ)awayt\left (1 - \dfrac {1}{\mu}\right ) awayt(1−μ1​)away

t(1+1μ)awayt\left (1 + \dfrac {1}{\mu}\right ) awayt(1+μ1​)away

t(1−1μ)nearert\left (1 - \dfrac {1}{\mu}\right ) nearert(1−μ1​)nearer

t(1+1μ)nearert\left (1 + \dfrac {1}{\mu}\right ) nearert(1+μ1​)nearer

Solution

Question 4

40 cm40\ cm40 cm

20 cm20\ cm20 cm

80 cm80\ cm80 cm

30 cm30\ cm30 cm

Solution

Question 5

0.25×10−10S0.25\times 10^{-10}S0.25×10−10S

1.67×10−7S1.67\times 10^{-7}S1.67×10−7S

2.5×10−10S2.5\times 10^{-10}S2.5×10−10S

1.167×10−10S1.167\times 10^{-10}S1.167×10−10S

Solution

Question 6

Telescope

Microscope

Gallilean telescope

Concave lens

Solution

Question 7

Inverted virtual image

Inverted real magnified image

Upright virtual image

Upright real magnified image

Solution

Question 8

sin−1(1μ)sin^{-1}\left ( \dfrac{1}{\mu } \right )sin−1(μ1​)

tan−1(1μ)tan^{-1}\left ( \dfrac{1}{\mu } \right )tan−1(μ1​)

sin−1(μ−1μ)sin^{-1}\left ( \dfrac{\mu -1}{\mu } \right )sin−1(μμ−1​)

cos−1(1μ)cos^{-1}\left ( \dfrac{1}{\mu } \right )cos−1(μ1​)

Solution

Question 9

None of these

Solution

Question 10

1.00 mm

1.50mm

2.50 mm

3.00 mm

Solution

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$15.5\ m$

$14.3\ m$

$13.5\ m$

$13.3\ m$

$t\left (1 - \dfrac {1}{\mu}\right ) away$

$t\left (1 + \dfrac {1}{\mu}\right ) away$

$t\left (1 - \dfrac {1}{\mu}\right ) nearer$

$t\left (1 + \dfrac {1}{\mu}\right ) nearer$

$40\ cm$

$20\ cm$

$80\ cm$

$30\ cm$

$0.25\times 10^{-10}S$

$1.67\times 10^{-7}S$

$2.5\times 10^{-10}S$

$1.167\times 10^{-10}S$

$sin^{-1}\left ( \dfrac{1}{\mu } \right )$

$tan^{-1}\left ( \dfrac{1}{\mu } \right )$

$sin^{-1}\left ( \dfrac{\mu -1}{\mu } \right )$

$cos^{-1}\left ( \dfrac{1}{\mu } \right )$