A 4 cm candle sits 15 cm in front of a converging lens with a focal length of 6.0 cm. Draw a ray diagram showing the image used and determine the image distance and magnification. Please help me answer!

Sure! Here's a step-by-step guide on how to draw a ray diagram and find the image distance and magnification for this scenario:

Step 1: Draw the principal axis
Draw a horizontal line to represent the principal axis. Label it with an arrow on both ends to indicate the direction of light.

Step 2: Place the lens
Draw a vertical line perpendicular to the principal axis to represent the converging lens. Label the lens with its focal length (6.0 cm) and the optical center (O).

Step 3: Place the object
Measure and mark a point on the principal axis at a distance of 15 cm in front of the lens. Label this point as the object (O') and draw an arrow above it to represent the candle.

Step 4: Ray 1 - Parallel Ray
Draw a straight line from the top of the candle parallel to the principal axis. The ray should pass through the lens and bend towards the focal point on the opposite side. Label this ray as Ray 1.

Step 5: Ray 2 - Focal Ray
Draw a straight line from the top of the candle through the focal point on the lens. The ray should be parallel to the principal axis after it passes through the lens. Label this ray as Ray 2.

Step 6: Ray 3 - Center Ray
Draw a straight line from the top of the candle passing through the center of the lens (O). This ray should not change direction since it is passing through the center of the lens. Label this ray as Ray 3.

Step 7: Determine the image
Where Ray 2 and Ray 3 intersect after passing through the lens, draw a dashed line perpendicular to the principal axis. This line represents the virtual image formed by the lens. Label the image as I' on the opposite side of the lens from the object.

Step 8: Image distance
Measure the distance between the lens and the image (I') along the principal axis. In this case, the image distance is 6.0 cm.

Step 9: Magnification
To determine the magnification, measure the height of the image (I') and the height of the object (O'). Divide the height of the image by the height of the object. The magnification in this case is -1.5, which means the image is inverted and smaller than the object.

Remember, distances above the principal axis should be positive (+) and distances below the principal axis should be negative (-).

I hope this helps you with your question! If you have any more doubts, feel free to ask!

To draw the ray diagram, we can follow these steps:

Step 1: Draw the principal axis (a horizontal line passing through the center of the lens).

Step 2: Place the converging lens on the principal axis with its focal point on the right side.

Step 3: Mark the object. In this case, draw an arrow (representing the candle) below the principal axis, 15 cm to the left of the lens.

Step 4: Draw three rays:

- Ray 1: Parallel to the principal axis that passes through the lens and then through the focal point on the right side.
- Ray 2: Passes through the center of the lens (it will remain undeviated).
- Ray 3: Passes through the focal point on the left side, and then through the lens parallel to the principal axis.

Step 5: Extend the rays until they intersect. The point where the rays converge will represent the image formed by the lens.

To determine the image distance and magnification, we can use the lens formula:

1/f = 1/v - 1/u,

where:
f = focal length of the lens (6.0 cm),
v = image distance, and
u = object distance (15 cm).

Substituting the values, we have:

1/6.0 = 1/v - 1/15.

Rearranging the equation:

1/v = 1/6.0 + 1/15.

Getting a common denominator:

1/v = (15 + 6.0) / (6.0 * 15).

Simplifying:

1/v = 21 / 90.

Multiplying both sides by v:

v = 90 / 21.

Calculating the value:

v ≈ 4.286 cm.

Now, to determine the magnification, we use the formula:

magnification = -v/u,

where
v = image distance (4.286 cm), and
u = object distance (15 cm).

Substituting the values:

magnification = -(4.286 cm) / (15 cm).

Calculating the value:

magnification ≈ -0.286.

Therefore, the image is formed approximately 4.286 cm from the lens and the magnification is approximately -0.286. The negative sign in the magnification indicates that the image is inverted compared to the object.