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3 years ago

7

Which of the following always occurs when a light ray reflects off a mirror? A.the speed of the light ray increases B. The direc

tion of the light ray stays the same C. The frequency of the light ray decreases as it reflects and loses energy to the mirror D. The angle at which the light ray strikes the mirror equals the angle at which it reflects

2 answers:

kakasveta [241]3 years ago

8 0

<span>The angle at which the light ray strikes the mirror equals the angle at which it reflects. Its the first law of reflection.</span>

Savatey [412]3 years ago

5 0

A, B, and C never happen. D always happens.

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Is the average speed of a vehicles vector or a scalar quantity? A) vectorB) scalar

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10 months ago

A concave mirror with a radius of curvature of 20 cm has a focal length of

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Answer:

A concave mirror has a radius of curvature of 20 cm. What is it's focal length? If an object is placed 15 cm in front of it, where would the image be formed? What is it's magnification?

The focal length is of 10 cm, object distance is 30 cm and magnification is -2.

Explanation:

Given:

A concave mirror:

Radius of curvature of the mirror, as C = 20 cm

Object distance in-front of the mirror = 15 cm

a.

Focal length:

Focal length is half of the radius of curvature.

Focal length of the mirror = $\frac{C}{2}$ = 10 cm

According to the sign convention we will put the mirror on (0,0) point, of the Cartesian coordinate open towards the negative x-axis.

Object and the focal length are also on the negative x-axis where focal length and image distance will be negative numerically.

b.

We have to find the object distance:

Formula to be use:

⇒ $\frac{1}{focal\ length}= \frac{1}{image\ distance} + \frac{1}{object\ distance}$

⇒ Plugging the values.

⇒ $\frac{1}{-10} =\frac{1}{image\ distance}+\frac{1}{-15}$

⇒ $\frac{1}{-10} -\frac{1}{-15}=\frac{1}{image\ distance}$

⇒ $\frac{1}{-10} + \frac{1}{15}=\frac{1}{image\ distance}$

⇒ $\frac{-3+2}{30} =\frac{1}{image\ distance}$

⇒ $\frac{-1}{30} =\frac{1}{image\ distance}$

⇒ $-30\ cm=image\ distance$

Image will be formed towards negative x-axis 30 cm away from the pole.

c.

Magnification (m) is the negative ratio of mage distance and object distance:

⇒ $m=-\frac{image\ distance}{object\ distance}$

⇒ $m=-\frac{(-30)}{(-15)}$

⇒ $m=-2$

The focal length of the concave mirror, is of 10 cm, object distance is 30 cm and magnification is -2.

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3 years ago

A paperweight consists of a 8.55-cm-thick plastic cube. Within the plastic a thin sheet of paper is embedded, parallel to opposi

ella [17]

Answer:

$\mu=1.5322$

Explanation:

Given:

Thickness of the paperweight cube, $x=8.55\ cm$

apparent depth from one side of the inbuilt paper in the plastic cube, $i=4.02\ cm$

apparent depth from the other side of the inbuilt paper in the plastic cube, $i'=1.56\ cm$

Now as we know that refractive index is given as:

$\rm \mu=\frac{real\ depth}{apparent\ depth}$

Let the real depth form first side of the slab be, $d$

Then the depth from the second side of the slab will be, $d'=x-d=8.55-d$

Since refractive index for an amorphous solid is an isotropic quantity so it remains same in all the direction for this plastic.

$\mu=\mu'$

$\frac{d}{i} =\frac{d'}{i'}$

$\frac{d}{4.02}=\frac{8.55-d}{1.56}$

$d=6.1597\ cm$

Now the refractive index:

$\mu=\frac{d}{i}$

$\mu=\frac{6.1596}{4.02}$

$\mu=1.5322$

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2 years ago