# Practical 4

Topics: Focal length, Optics, Lens Pages: 6 (1788 words) Published: February 2, 2013
Practical 4
Title: To study the magnification of a real image by a convex lens. Objective: To determine the focal length of a convex lens.
Apparatus and Materials:
1. Light box
2. Convex lens
3. Plasticine
4. Meter rule
5. Screen
6. Short transparent ruler
Setup:
1. Set up the apparatus as shown in Figure 4-1.

Figure 4-1
Theory:
From the lens equation:

Where:
p = object distance
q = image distance

Linear magnification,

Procedure:
1. The apparatus was set up as in Figure 4-1.
2. The light source was switched on.
3. A length of “1cm” on the scale of the transparent ruler was chose as the object, therefore object size, y = 1cm. 4. A value for the object distance, p was set. The image distance was adjusted until a sharp image was obtained on the screen. 5. The image distance, q was measured.

6. The length, y’ or size of the “1cm” image was measured. 7. The magnitude of the linear magnification, m was found for each image where .
8. The object distance was varied, steps 4 to 7 was repeated, and seven (7) sets of readings of p, q, y’ and m was obtained. 9. The readings of p, q, y’ and m was tabulated.
10. A graph of m against q was plotted.
11. The gradient of the graph was determined.
12. The focal length, f of the lens was calculated.

Data:
Size of object, y = ( 2.60 ± 0.10 ) cm
Object distance,p ± 0.10cm| Image distance,q ± 0.10 cm| Size of image,y' ± 0.10 cm| Magnification,m = y’/y| 15.0| 30.0| 5.20| 2.00|
20.0| 20.0| 2.70| 1.00|
25.0| 16.7| 1.70| 0.70|
30.0| 15.0| 1.30| 0.50|
35.0| 14.0| 1.00| 0.40|
40.0| 13.3| 0.80| 0.30|
45.0| 12.9| 0.60| 0.20|

Calculation:

Mean of image distance, q=30.0+20.0+16.70+15.0+14.0+13.3+12.97 = 121970
= 17.41 cm #

Mean of magnification, m, =2.00+1.00+0.70+0.50+0.40+0.30+0.207 = 5170
=0.73 cm #

Centroid = ( 17.41 , 0.73) #

Data Analysis:
| Linear Least Squares Fits| | | | |
| | | | | | | | |
| x| y| xy| x^2| Sx| Sx + c| y - (Sx + c)| [y - (Sx + c)]^2| 1| 30.000| 2.000| 60.0000| 900.0000| 3.0692| 2.0162| -0.0162| 0.0003| 2| 20.000| 1.000| 20.0000| 400.0000| 2.0461| 0.9931| 0.0069| 0.0000| 3| 16.700| 0.700| 11.6900| 278.8900| 1.7085| 0.6555| 0.0445| 0.0020| 4| 15.000| 0.500| 7.5000| 225.0000| 1.5346| 0.4816| 0.0184| 0.0003| 5| 14.000| 0.400| 5.6000| 196.0000| 1.4323| 0.3793| 0.0207| 0.0004| 6| 13.300| 0.300| 3.9900| 176.8900| 1.3607| 0.3077| -0.0077| 0.0001| 7| 12.900| 0.200| 2.5800| 166.4100| 1.3198| 0.2667| -0.0667| 0.0045| 8|  |  |  |  |  |  |  |  |

9|  |  |  |  |  |  |  |  |
10|  |  |  |  |  |  |  |  |
11|  |  |  |  |  |  |  |  |
12|  |  |  |  |  |  |  |  |
| 121.900| 5.100| 111.3600| 2343.1900|  |  |  | 0.0076| | | | | | | | | |
| n =| 7| | |  =| 0.0389| | |
| | | | | | | | |
| S =| 0.1023| | | (S)=| 0.0026| | |
| | | | | | | | |
| c =| -1.0530| | | (c)=| 0.0480| | |
f=1S= 10.1023 =9.775 cm | σ f = σ ( S )S ×f = 0.0260.1023 × 9.775 cm =0.2484 cm| fexp=9.775 ±0.25 cm| Known focal length,fk=10.00 cm| % error= f exp- fkfk ×100% = 9.775-10.0010.00 ×100% =2.25 %| Percentage Error=2.25 %|

Results :
The focal length of the convex lens was found to be f = ( 9.775 ± 0.25 ) cm with a percentage error of 2.25 % Discussion:
A convex lens (converging lens) is a circular glass plate convex on both surfaces. The non-uniform thickness causes bending of light towards the principle axis. In particular, a convex lens...