Concave Mirror and Thin Lens Equation Quiz

Concave Mirror and Thin Lens Equation Quiz

Assessment

Interactive Video

Physics

9th - 10th Grade

Hard

Created by

Jennifer Brown

FREE Resource

10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the primary purpose of using the thin lens equation in this video?

To determine the color of the image

To calculate the image size, distance, orientation, and type

To find the weight of the object

To measure the temperature of the mirror

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Why is it important to note the sign of the focal length for concave mirrors?

It alters the shape of the object

It changes the material of the mirror

It affects the calculation of image distance

It determines the color of the image

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the focal length of the concave mirror used in the problem?

7 cm

5 cm

4 cm

6 cm

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the significance of a negative image distance in the context of this problem?

The image is smaller than the object

The image is larger than the object

The image is on the other side of the mirror

The image is in front of the mirror

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How is the image distance calculated using the thin lens equation?

By adding the object distance and focal length

By subtracting the object distance from the focal length

By multiplying the object distance and focal length

By taking the reciprocal of the sum of reciprocals of object distance and focal length

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What does a positive magnification indicate about the image?

The image is inverted

The image is on the other side of the mirror

The image is smaller than the object

The image is upright

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How is the magnification of the image calculated?

By adding the object height and image height

By multiplying the object distance by the image distance

By dividing the object height by the image height

By dividing the negative image distance by the object distance

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