problem 7

16
Problem 7 To solve this problem you needed to know that: 1. square has two pairs of parallel sides that intersect at 90° angle; 2. real image is formed by converging lens if the object´s distance is greater than focal length (from Newton´s equation: f 2 =x 1 x 2 ); 3. rays intersecting first focal point will be parallel to optical axis after passing through the lens; 4. two parallel rays at an angle relative to optical axis will intersect at point lying in the second focal plane of the lens – this point is found by using the above (hint 3.) and; 5. basic trigonometry

Upload: zoltin

Post on 20-Mar-2016

23 views

Category:

Documents


2 download

DESCRIPTION

Problem 7. To solve this problem you needed to know that: square has two pairs of parallel sides that intersect at 90° angle; real image is formed by converging lens if the object´s distance is greater than focal length (from Newton´s equation: f 2 =x 1 x 2 ); - PowerPoint PPT Presentation

TRANSCRIPT

Page 1: Problem 7

Problem 7To solve this problem you needed to know that:

1. square has two pairs of parallel sides that intersect at 90° angle;2. real image is formed by converging lens if the object´s distance is

greater than focal length (from Newton´s equation: f2=x1x2);

3. rays intersecting first focal point will be parallel to optical axis after passing through the lens;

4. two parallel rays at an angle relative to optical axis will intersect at point lying in the second focal plane of the lens – this point is found by using the above (hint 3.) and;

5. basic trigonometry

Page 2: Problem 7

• Opposite sides of this quadrilateral are formed by initialy parallel sides of the square.

Page 3: Problem 7

Points at which opposite sides intersect should be lying in the second focal plane of the lens

Page 4: Problem 7

For ideal thin lens, focal planes are not curved, so we simply connect points F´ and F´´

Page 5: Problem 7

To find the solution, let us assume that the focal lenght is i.e. 3 cm

Page 6: Problem 7

Using the hints 3. and 4. in reverse, two rays (shown in yellow) should be parallel to sides of the square, and therefore meet at 90°angle. Also, they intersect at the first focal point

Page 7: Problem 7
Page 8: Problem 7

To find focal length and optical axis offset (a1 or a2) we mark angles of rays relative to optical axis (sides of the square!)

Page 9: Problem 7

Let us mark the distance F´F´´ as d. This distance is a indenpendent from the focal lenght and offset so, by using trigonometry we have:

(1)

Page 10: Problem 7

Using hints 3. and 4. again we can find the position of the “square”

Page 11: Problem 7

Let us mark the “square” side as s´ and translate it to form a triangle with sides s´(red), d´(brown) and angles equal to α,β and 90°.We repeat the same for side s´´. This gives us:

(2)

Page 12: Problem 7

To find distances d´and d´´ , we will use angles Δ, Δ´, δ, δ´. Note that these angles are indenpendent from the focal length and offset, so we have:

(3)

(4)

Page 13: Problem 7

Using all four equation, we have:

Page 14: Problem 7

From this picture; we can see that angles Δ, Δ´, δ, δ´ are indeed indenpendent. Their values are:

Using these data, we obtain:

Page 15: Problem 7

The complete solution occupies only one page

Page 16: Problem 7

To avoid taking the derivatives, uncertanities are found by directly inputing the data, and calculating the final uncertanity.