Experiment 403: Refraction from a Spherical Surface: Thin Lens Mapua Institute of Technology I. Abstract Refraction is the bending of light ray as it hits a material of different optical density. Transparent materials like lenses can refract parallel ray of light and can produce an image. If the middle part of lens is thicker, it is called a converging lens or a convex lens. But if the middle part is thinner, the lens is called diverging lens or concave lens. The main effect of lenses is to refract light passing through it. It is made up of transparent substances that are bounded by two surfaces of regular form. However, it is necessary that the index of refraction of the material should be higher than that of the surrounding medium. Generally, lenses are classified as either converging or diverging. Converging lenses are thicker in the middle than they are in the edge. Conversely, diverging lenses are thicker at the edges than at the middle, so that, the incident waves to the surface parallel to lens axis diverge from a point in front of the lens (principal focus). II. Objectives For Experiment 403 entitled as “Refraction from a Spherical Surface: Thin Lens”, one can state that the study seeks (i) to determine the focal length of a convex lens using (a) different locations of the object and (b) the graphical method. III. Results and Discussion This experiment tackles the concept of thin lens and refraction of a light from a spherical surface. Refraction can be described as a wave bent when it passes a certain medium with different optical density. There are many types of lens but in this experiment we mainly focused on the properties of a converging lens. This lens has the ability to change the shape of a certain wave that pass through it. In the first part, we determine the focal length of the two lens using an object at infinity. We used the light passing through the window of our room as our object at infinity, we adjusted the lens until a sharp image of the window is formed. We have observed that the image created is real and inverted. In the second part of our experiment, we determine the focal length using an object at a finite distance. We used the light source, convex lens and image screen for this part. The screen is placed 1 meter away from the light source. The lens are the ones that will be moved until the object projected from the screen is sharp. We have observed in the lens 1 that the image projected is larger than the original image and inverted but
when the lens is closer from the image screen, the image projected becomes smaller. Therefore if the object distance is greater than the image distance the size of the image projected decreases but when the image distance is greater than the object distance the image projected is larger. The object and image distance can an interchanging values which gives a point of conjunction but still it will yield a similar value for f based on the data we gathered. In the last part of the experiment, we are tasked to determine the focal length using the graphical technique. It can be observed in the graph that as 1/s decreases, 1/s’ increases therefore 1/s and 1/s’ has an inversely proportional relationship. The focal length is obtained by using the reciprocal of the intercepts. If we compare the average value of the focal length gathered by our group to its actual value, it has similar values because we obtained a percentage error of 0.31%. Also by observing, position 1 and 2 that are created by the line coincides with one another which proves the theory that image and object distances are really interchangeable. Lastly, the magnification of the projected image indicates the orientation of an object. Having a positive magnification indicates that the orientation of the image is upright, while if the magnification value is negative it means that the object is inverted. When the magnification value is less than one, the image it projects is smaller but when the magnification value is greater than one, the image it projects is larger. By using these information and the data we gathered we can say that the image projected in this part of our experiment has an orientation opposite from the object, because our values are negative. IV. Conclusion I can therefore conclude that, refraction is a phenomenon that occurs when the light ray is bent when it hits a certain object. Lens is a device made of transparent material that refracts the light rays going through it. In this experiment we mainly used convex lens, it has 2 focal points which is one on each sides, because of that the values of the image and object distance can be interchanged and it can form a real inverted image or a virtual erected image which depends on the values of the distances of the focal length and object. If the object is outside of the image’s focal length it can be concluded that the image formed is real and inverted, but if the object is inside the focal length the image formed is virtual and upright. We can differentiate the concave and convex lens by the image it projects. Convex lens projects a real image while a concave lens projects a virtual image. Magnification can be described as the ratio of an image size. Magnification has different properties, when its value is negative it indicates that the image is inverted. If the value of the magnification is positive it can be said that the orientation is upright. If the value is greater than 1 we can
say that the image magnified is enlarged but if the value is less than 1 the image size is diminished. Therefore we have proved that image and object distances are interchangeable which we can conclude that it has points of conjunction that will yield the same value for the focus.
Sample Computation: Part A: Determination of Focal Length using an Object at Infinity Solution: Lens 1 (Trial 1) 1 f= =9.9 cm 1 1 + ∞ 9.9 error =
10−10.05 x 100 =0.5 10
Part B: Determination of Focal Length using an Object at Finite Distance Solution: Lens 1 (Position 1) 1 f= =9.8 cm 1 1 + 11 90 error =
10−9.795 x 100 =2.05 10
Part C: Determination of Focal Length using Graphical Technique Solution: Lens 2 (Position 1) 1 1 1 1 = =0.033898 c m−1 = =0.014184 c m−1 100-cm Gap: s 29.5 s ' 70.5 90-cm Gap:
1 1 −1 = =0.030675 c m s 32.6
1 1 −1 = =0.017422c m s ' 57.4
85-cm Gap:
1 1 = =0.025907 c m−1 s 38.6
1 1 = =0.021552 c m−1 s ' 46.4
Equation:
y=−0.9177 x+ 0.0454
−1
x-intercept: 0.049472 c m 1 x−∫ ¿= =20.21366 cm 0.049472 1 ¿
Focal Length:
−1
y-intercept: 0.0454 c m 1 y−∫ ¿= =22.02643 cm 0.0454 1 ¿
Focal Length:
Focal Length: Ave Focal Length= error =
20.21366+22.02643 =21.2004 cm 2
20−21.2004 x 100 =5.60022 20
Magnification: '
100-cm Gap: difference=
m=
−s 70.5 = =−2.39 s 29.5
2.39−2.425 x 100 =1.45 2.39+2.425 2
m=
hi 9.7 = =2.425 ho 4
