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larisa86
6 days ago
7

A convex mirror with a focal length of 0.25 m forms a 0.080 m tall image of an automobile at a distance of 0.24 m behind the mir

ror. What is the magnification of the image? Where is the car located, and what is its height? Is the image real or virtual? Is the image upright or inverted? Draw a ray diagram to show where the image forms and how large it is with respect to the object

Physics
2 answers:
ValentinkaMS [1.1K]6 days ago
6 0

Answer:

The object measures 6 m in distance and 2 m in height.

It creates a virtual image that is upright.

Explanation:

Provided data includes:

Focal length = 0.25 m

Image height = 0.080 m

Image distance = 0.24 m

We are to determine the object's distance.

Using the lens formula:

\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}

Substituting values into the formula:

\dfrac{1}{0.24}=\dfrac{1}{0.25}+\dfrac{1}{u}

\dfrac{1}{u}=\dfrac{1}{0.24}-\dfrac{1}{0.25}

\dfrac{1}{u}=\dfrac{1}{6}

u=6\ m

We also need to calculate magnification:

Applying the magnification formula:

m=-\dfrac{v}{u}

Substituting values into this formula:

m=-\dfrac{0.24}{-6}

m=0.04

Next, we need to find the height of the object:

Using the magnification formula once more:

m=\dfrac{h'}{h}

h=\dfrac{0.080}{0.04}

h=2\ m

A convex mirror generates a virtual and upright image on its backside.

Consequently, the object is at a distance of 6 m and has a height of 2 m.

The image formed is virtual and upright.

ValentinkaMS [1.1K]6 days ago
3 0

Answer:

Object distance = 6 m

Object height = 2 m

Explanation:

Reasoning process:

Provided that,

Focal length = 0.25 m

Image length = 0.080 m

Image distance = 0.24 m

We must determine the object's distance.

Thus, applying the lens formula:

\frac{1}{u} = \frac{1}{f} + \frac{1}{u}

\frac{1}{u} = \frac{1}{6}

Calculation yields u = 6

The magnification can be found as:

m = -0.24 / -6

= 0.04

The height = 2 m

The illustration shows an upright image positioned behind the mirror.

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Answer:

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We have a set of statements to evaluate for correctness. The most effective approach is to examine the problem in detail.

Using the equation for vertical launch, we acknowledge that the positive direction signifies downward movement.

Stone 1

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For t > = to, the expression (2t/to-1) yields a value greater than 1, indicating that the distance expands as time progresses.

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Note that we don’t need to be concerned with the signs here, as all entities are moving toward each other. If something was moving away, a negative sign would apply, but that is not relevant to this scenario.

Where,
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