991_PartUniversity Physics Solution

# 991_PartUniversity Physics Solution - Geometric Optics...

This preview shows pages 1–3. Sign up to view the full content.

Geometric Optics 34-29 Then 1.33 1.55 1.55 1.33 20.0 cm 2.50 cm s += gives 72.1 cm s ′ = − . The image is 72.1 cm to the left of the surface vertex. EVALUATE: With the rod in air the image is real and with the rod in water the image is virtual. 34.96. IDENTIFY: Apply 11 1 s sf to each lens. The image formed by the first lens serves as the object for the second lens. The focal length of the lens combination is defined by 12 111 s + = . In part (b) use 11 1 (1 ) n f RR ⎛⎞ =− ⎜⎟ ⎝⎠ to calculate f for the meniscus lens and for the 4 CCl , treated as a thin lens. SET UP: With two lenses of different focal length in contact, the image distance from the first lens becomes exactly minus the object distance for the second lens. EXECUTE: (a) 11 1 1 11 111 111 s sfs +=⇒=− ′′ and 22 2 2 11 1 1 11 11 . s ss f s f += += − += But overall for the lens system, 21 . s fff +=⇒=+ (b) With carbon tetrachloride sitting in a meniscus lens, we have two lenses in contact. All we need in order to calculate the system&s focal length is calculate the individual focal lengths, and then use the formula from part (a). For the meniscus lens 1 m1 2 1 1 1 ( ) (0.55) 0.061 cm 4.50 cm 9.00 cm ba nn fR R = = and m 16.4 cm f = . For the 1 4 w1 2 1 1 1 CCl : ( ) (0.46) 0.051 cm 9.00 cm R = − = and w 19.6 cm f = . 1 wm 0.112 cm ff f =+= and 8.93 cm f = . EVALUATE: f f f f f = + , so f for the combination is less than either 1 f or 2 f . 34.97. IDENTIFY: Apply Eq.(34.11) with R →∞ to the refraction at each surface. For refraction at the first surface the point P serves as a virtual object. The image formed by the first refraction serves as the object for the second refraction. SET UP: The glass plate and the two points are shown in Figure 37.97. plane faces means R and 0 ab + = b a n s s n ′=− Figure 34.97 EXECUTE: refraction at the first (left-hand) surface of the piece of glass: The rays converging toward point P constitute a virtual object for this surface, so 14.4 cm. s 1.00, 1.60. == 1.60 ( 14.4 cm) 23.0 cm 1.00 s =+ This image is 23.0 cm to the right of the first surface so is a distance 23.0 cm t to the right of the second surface. This image serves as a virtual object for the second surface. refraction at the second (right-hand) surface of the piece of glass: The image is at P so 14.4 cm 0.30 cm 14.7 cm . s tt =+− (23.0 cm ); 1.60; 1.00 st n n = = b a n s s n gives 1.00 14.7 cm ( [23.0 cm ]). 1.60 −=− 14.7 cm 14.4 cm 0.625 . − =+ 0.375 0.30 cm t = and 0.80 cm t = EVALUATE: The overall effect of the piece of glass is to diverge the rays and move their convergence point to the right. For a real object, refraction at a plane surface always produces a virtual image, but with a virtual object the image can be real.

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

View Full Document
34-30 Chapter 34 34.98. IDENTIFY: Apply the two equations 11 1 22 2 and abba bccb nnnn ss R += ′′ . SET UP: liq ac nn n == , b nn = , and 12 s s ′ = − . EXECUTE: (a) liq liq liq liq 1 2 and n n n n ss R R −− + = .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 5

991_PartUniversity Physics Solution - Geometric Optics...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online