Tag: conditions for a light ray to pass undeviated

Quizzes (0) Questions (20)

Questions Related to conditions for a light ray to pass undeviated

Multiple choice conditions for a light ray to pass undeviated refraction of light at plane surfaces physics

A 5 cm thick glass slab is silvered at one of the surface to form a plane mirror. A point object is placed at a distance of $20cm $ from the unsilvered face. The image distance from the unsilvered face will be [R.I. of glass =1.5]

  1. 26.67cm

  2. $\frac {70}{3}$

  3. 20cm

  4. 25cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The object is at 20 cm from the unsilvered face. The light travels through 5 cm of glass (mu=1.5) to reach the mirror. The apparent distance of the object from the mirror is 5/1.5 + 20 = 3.33 + 20 = 23.33 cm. The mirror forms an image at 23.33 cm behind the mirror. The light then travels back through the glass, where the image appears at an apparent distance of 23.33 * 1.5 = 35 cm from the mirror surface. The total distance from the unsilvered face is 35 + 5 = 40 cm? Wait, recalculate: The mirror is at depth 5 cm. Object is at 20 cm from surface. Distance to mirror = 20 + 5/1.5 = 20 + 3.33 = 23.33. Image is at 23.33 behind mirror. Total distance from surface = 23.33 + 5/1.5 = 23.33 + 3.33 = 26.66 cm.

Multiple choice conditions for a light ray to pass undeviated refraction of light at plane surfaces physics

Light starting from a medium of refractive
index $  \mu  $ undergoes refraction into a medium of index $  \mu  $. If I and r stand for angle of incidence are refraction respectively :

  1. $

    \frac{\sin i}{\sin r}=\frac{\mu}{\mu^{\prime}}

    $

  2. $

    \frac{\sin i}{\sin r}=\frac{\mu^{\prime}}{\mu}

    $

  3. $

    \frac{\cos i}{\cos r}=\frac{\mu^{\prime}}{\mu}

    $

  4. $

    \frac{\sin i}{\sin r}=1 / \mu \mu^{\prime}

    $
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice conditions for a light ray to pass undeviated refraction of light at plane surfaces physics

When a light ray is refracted from one medium into another the wavelength change from $4500{ A }^{ 0 }$ to $3000{ A }^{ 0 }$ the critial angle for a ray from second medium to first medium is 

  1. ${ sin }^{ -1 }\left( \dfrac { 2 }{ 3 } \right) $

  2. ${ cos }^{ -1 }\left( \dfrac { 2 }{ 3 } \right) $

  3. ${ tan }^{ -1 }\left( \dfrac { 3 }{ 2 } \right) $

  4. ${ tan }^{ -1 }\left( \dfrac { 2 }{ \sqrt { 5 } } \right) $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Refractive index mu = lambda1 / lambda2 = 4500 / 3000 = 1.5 = 3/2. The critical angle C is given by sin(C) = 1 / mu = 1 / 1.5 = 2/3. Thus, C = sin^-1(2/3).

Multiple choice conditions for a light ray to pass undeviated refraction of light at plane surfaces physics

Which of the following statements is false?

  1. A plane mirror produce a magnification of +1

  2. Focal length of a plane mirror is infinite

  3. For man of height h, to see his own complete image , a mirror of height at least equal to  h is required

  4. For a ray of light, incidenting normally on a plane mirror, the angle of reflection is $ 180^0 $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

To see a full image in a plane mirror, the mirror height must be at least half the person's height (h/2), not equal to h. Thus, statement C is false.

Multiple choice conditions for a light ray to pass undeviated refraction of light at plane surfaces physics

A wave has velocity $u$ in medium $P$ and velocity $2u$ in medium $Q$. If the wave is incident in medium $P$ at an angle of $30^\circ$ then the angle of refraction will be

  1. $30^\circ$

  2. $45^\circ$

  3. $60^\circ$

  4. $90^\circ$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using Snell's Law: sin(i) / sin(r) = v1 / v2. Given i = 30 degrees, v1 = u, v2 = 2u. sin(30) / sin(r) = u / 2u = 1/2. sin(r) = 2 * sin(30) = 2 * 0.5 = 1. Therefore, r = 90 degrees.

Multiple choice conditions for a light ray to pass undeviated refraction of light at plane surfaces physics

What is the angle of incidence if a ray of light is not deviated when it enters a block of glass?

  1. $30^{\circ}$

  2. <span>$60^{\circ}$</span>

  3. <span>$90^{\circ}$</span>

  4. <span>$0^{\circ}$</span>
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The angle of incidence if a ray of light is not deviated when it enters a block of glass :

The angle of incidence is equal to angle of refraction only if the angle the  deviation is zero. This is practically impossible since, when light enters a medium from another medium it has to undergo refraction. So angle of minimum deviation is never zero but it can be a very small value. And angle of incidence is never equal to angle of refraction, if the light undergoes minimum deviation. But the angle of incidence is equal to the angle of emergence.

Multiple choice conditions for a light ray to pass undeviated refraction of light at plane surfaces physics

An under water swimmer looks upward at an unobstructed overcast sky. The vertex angle does the sky appear to subtend at the eye of swimmer is (critical angle for water air interface is  $C$).

  1. $C$

  2. $C/2$

  3. $2 C$

  4. $3 C$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A swimmer looking up sees the sky through a cone of light defined by the critical angle C. The total angle subtended at the eye is 2 * C.

Multiple choice conditions for a light ray to pass undeviated refraction of light at plane surfaces physics

A mark is made on the surface of a glass sphere of diameter 10 cm and refractive index 1.5 . it its viewed through the glass from a potion directly opposite . the distance of the image of the mark from the centre of the sphere will be 

  1. 20 cm

  2. 17.5 cm

  3. 15 cm

  4. 22.5 cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the refraction formula at a spherical surface: (mu2 / v) - (mu1 / u) = (mu2 - mu1) / R. Here, the object is on the surface, so u = -10 cm (diameter). mu1 = 1.5, mu2 = 1, R = -5 cm. (1 / v) - (1.5 / -10) = (1 - 1.5) / -5 => 1/v + 0.15 = 0.1 => 1/v = -0.05 => v = -20 cm. The image is 20 cm from the pole, which is 20 cm from the center.