{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chapter 24

# Chapter 24 - 24.1 Model Each lens is a thin lens The image...

This preview shows pages 1–4. Sign up to view the full content.

24.1. Model: Each lens is a thin lens. The image of the first lens is the object for the second lens. Visualize: The figure shows the two lenses and a ray-tracing diagram. The ray-tracing shows that the lens combination will produce a real, inverted image behind the second lens. Solve: (a) From the ray-tracing diagram, we find that the image is 50 cm from the second lens and the height of the final image is 4.5 cm. (b) s 1 = 15 cm is the object distance of the first lens. Its image, which is a virtual image, is found from the thin- lens equation: 11 1 111 1 1 5 40 cm 15 cm 120 cm sf s =−= = 1 24 cm s ′=− The magnification of the first lens is ( ) 1 1 1 24 cm 1.6 15 cm s m s =− = The image of the first lens is now the object for the second lens. The object distance is s 2 = 24 cm + 10 cm = 34 cm. A second application of the thin-lens equation yields: 22 2 111 1 1 20 cm 34 cm s 2 680 cm 48.6 cm 14 s == The magnification of the second lens is 2 2 2 48.6 cm 1.429 34 cm s m s The combined magnification is ( )( ) 12 1.6 1.429 2.286 mm m == − = . The height of the final image is (2.286)(2.0 cm) = 4.57 cm. These calculated values are in agreement with those found in part (a).

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
24.2. Model: Each lens is a thin lens. The image of the first lens is the object for the second lens. Visualize: The figure shows the two lenses and a ray-tracing diagram. The ray tracing shows that the lens combination will produce a virtual, inverted image in front of the second lens. Solve: (a) From the ray-tracing diagram, we find that the image is 20 cm in front of the second lens and the height of the final image is 2.0 cm. (b) s 1 = 60 cm is the object distance of the first lens. Its image, which is a real image, is found from the thin-lens equation: 11 111 1 1 1 40 cm 60 cm 120 cm s sfs =−= = 1 120 cm s ′ = The magnification of the first lens is 1 1 1 120 cm 2 60 cm s m s = −= = The image of the first lens is now the object for the second lens. The object distance is s 2 = 160 cm 120 cm = 40 cm. A second application of the thin-lens equation yields: 22 2 1 1 1 40 cm 40 cm ss f =− + = + +− 2 20 cm s ′=− The magnification of the second lens is 2 2 2 20 cm 0.5 40 cm s m s = = + The overall magnification is ( )( ) 12 20 . 5 1 . 0 mm m == = . The height of the final image is ( + 1.0)(2.0 cm) = 2.0 cm. The image is inverted because m has a negative sign. These calculated values are in agreement with those found in part (a).
24.3. Model: Each lens is a thin lens. The image of the first lens is the object for the second lens. Visualize: The figure shows the two lenses and a ray-tracing diagram. The ray tracing shows that the lens combination will produce a real, upright image behind the second lens. Solve: (a) From the ray-tracing diagram, we find that the image is 10 cm behind the second lens and the height of the final image is 2 cm.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 56

Chapter 24 - 24.1 Model Each lens is a thin lens The image...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online