3404 Thin Lenses

(6) What image does an object more than twice the distance of the focal length from a converging mirror produce?

Solution

Real, invered and reduced.

(7) The near point for a far-sighted person is at 100cm. What focal length will allow the person to clearly see an object 25cm away?

Solution

33cm

Exercises

27, 28, 30, 32

34.27 An insect 4.25 mm tall is placed 20.0 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the right surface has a radius of curvature of magnitude 12.3 cm , and the index of refraction of the lens material is 1.70. (a) Calculate the location of the image this lens forms of the insect. (b) Calculate the size of the image.

Solution

1. The lensis a converging lens. The focal length can be resolved by lensmaker's formula.

1f=(n−1)(1R1−1R2)R1=12.3cmR2=∞⇒f=17.6cm\begin{aligned} \frac{1}{f} &= (n-1)(\frac{1}{R_1} - \frac{1}{R_2})\\ R_1 &= 12.3 \text{cm} \quad R_2 = \infty\\ \To f &= 17.6\text{cm} \end{aligned}

2. The image distance from the lens.

1f=1I+1OO=20⇒I=145cm\begin{aligned} \frac{1}{f} &= \frac{1}{I} + \frac{1}{O}\\ O &= 20\\ \To I &= 145 \text{cm} \end{aligned}

3. The image size is

hi=∣m∣hom=−IOho=4.25mm⇒hi=30.75mm\begin{aligned} h_i &= |m|h_o\\ m &= -\frac{I}{O} \quad h_o= 4.25\text{mm}\\ \To h_i &= 30.75\text{mm} \end{aligned}

4. If the lensis reversed, the result is the same.

34.28 A lens forms an image of an object. The object is 16.0 cm from the lens. The image is 12.0 cm from the lens on the same side as the object.
(a) What is the focal length of the lens?
(b) Is the lens converging or diverging?
(c) If the object is 8.50 mm tall, how tall is the image?
(d) Is it erect or inverted?

Solution

a. f=−48cmf=-48\text{cm}
b. Diverging
c. 8.5×1216=0.638cm8.5 \times \frac{12}{16} = 0.638\text{cm}
d. Erect virtual image.

34.30 A converging lens with a focal length of 70.0 cm forms an image of a 3.20-cm-tall real object that is to the left of the lens. The image is 4.50 cm tall and inverted.
(a) Where is the image located in relation to the lens?
(b) Is the image real or virtual?
(c) How much to the left of the lens is the object?
(d) How far from the lens is the image? Follow the sign rules.

Solution

The image is real and located at the right of the lens.

1f=1I+1OIO=4.53.2f=70cm⇒I=168cmO=120cm\begin{aligned} \frac{1}{f} &= \frac{1}{I} + \frac{1}{O}\\ \frac{I}{O} &= \frac{4.5}{3.2}\\ f &= 70\text{cm}\\ \To I &= 168\text{cm} \quad O=120\text{cm} \end{aligned}

34.32 A photographic slide is to the left of a lens. The lens projects an image of the slide onto a wall 6.00 m to the right of the slide. The image is 78.0 times the size of the slide.
(a) How far is the slide from the lens?
(b) Is the image erect or inverted?
(c) What is the focal length of the lens?
(d) Is the lens converging or diverging?

Solution

(a) The object distance is 678m=0.077m\frac{6}{78} \text{m} = 0.077\text{m}.
(b) The images is real and inverted.
(c) The focal length is f=(16+786)−1=0.076mf = (\frac{1}{6} + \frac{78}{6})^{-1} = 0.076\text{m}.
(d) THe lens is a converging lens.

Quiz

(11) Two lens combination is separated by 50cm. The focal length is 10cm, and the focal length of the second lens is 15cm. What is the resultant image position (measured from lens 2) and the magnification?

Solution

60cm, +6