Optical Lens and Sensor Setup: Detecting Magnified Images

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The discussion revolves around the use of a lens in conjunction with a light sensor for improved detection of an object. The primary inquiry is whether the lens magnifies the object for easier detection and if a screen is necessary for the sensor to view a real image formed by the lens. It is clarified that a light sensor can function as the screen itself, provided it is positioned correctly to capture the image formed by the lens. The conversation emphasizes the importance of the lens's focal length and the relative positioning of the object, lens, and sensor to optimize light collection. A convex lens is highlighted for its ability to converge light, increasing intensity on the sensor compared to direct detection without a lens. Calculations in ray optics are suggested to determine optimal placements for effective sensing.
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im in a slight fix.
i am presently using a light sensor. i was asked to use a lens along with it recently. is it because the obect i am going to sense is magnified by the lens so that sensor can detect it easily?

but, let's take the situation where a real image is formed because of the lens. Dont we need a screen in this case to project the image, for the sensor to view it? or is it that the sensor doesn't need a projected image.


lets imagine my sensor is a transmitter-detector led pair at an acute angle.
can a light sensor detect the converging(or diverging) rays from the lens, or do i require a proper projected image for sensing.

waiting for a reply.
 
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kdkdkd said:
lets imagine my sensor is a transmitter-detector led pair at an acute angle.
I am not quite sure what you mean by a detector LED. You can use an LDR as detector.
kdkdkd said:
i was asked to use a lens along with it recently. is it because the obect i am going to sense is magnified by the lens so that sensor can detect it easily?
Trivially, a convex lens is a converging lens in all but one positions of the object. This means that even if your source gives light that is diverged and hence has low intensity per unit area, the use of a convex lens will increase the intensity to some extent. But this would highly depend on the position of your source and the relative position of the detector wrt the source. If the position of your source is fixed, it will be easier. Do some calculations in ray optics to find where the rays will converge,and position your detector accordingly. You can use a laser for your source, thereby eliminating the need of a lens. Further calculations will depend on where you're positioning the source, how large your detector is, and what the separation between the source and detector.
kdkdkd said:
but, let's take the situation where a real image is formed because of the lens. Dont we need a screen in this case to project the image, for the sensor to view it? or is it that the sensor doesn't need a projected image.
That is not the case. I recommend you to read up something on ray optics for lens to better understand how things work.
 
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kdkdkd said:
lets take the situation where a real image is formed because of the lens. Dont we need a screen in this case to project the image, for the sensor to view it? or is it that the sensor doesn't need a projected image.
The light sensor IS the screen. Position the lens to put the image on the sensor.
kdkdkd said:
is it because the obect i am going to sense is magnified by the lens so that sensor can detect it easily?
A convex (positive) lens collects the light that shines on it and tends to aim it to a point. If you get the spacing between the object, the lens, and the sensor correct and also the focal length of the lens correct, the light collected by the lens will come to a point at the sensor. Since the lens is larger than the sensor, this results in more light on the sensor than the sensor by itself could collect.

Cheers,
Tom
 
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