You wish to project the of a slide on a screen 9.00 m from the lens of a slide projector. (b) If the dimensions of the picture on a 35-mm color slide are 24 mm * 36 mm, what is the minimum size of the projector screen required to accommodate the ?

Video transcript

Hi, everyone. In this practice problem, we're being asked to calculate or determine the minimum dimensions of a whiteboard to fit in the image dimensions. We'll have a science teacher desire to use a lens to project the screen of her smartphone onto a whiteboard three m away from the phone. The phone is placed 30 centimeters to the left of the lens and the screen of the smartphone has a width of five centimeter and a length of 10 centimeter. We're being asked to determine the minimum dimensions of the whiteboard necessary to fit the image dimensions. The options given for the minimum dimensions of the whiteboard are a width equals to 0.25 m and length equals to 0.75 m B width equals to 0.25 m and length equals to 0.5 m C width equals to 0.5 m and length equals to one m. And lastly D width equals to 0.75 m and length equals to 1.5 m. So in this practice problem, the object length distance or s will be equals to 0.3 m. While the image lens distance or S prime will be equals to 3.0 m. Therefore, the magnification factor M of the phone screen can be calculated so that M which can be obtained by um taking the negative of S prime divided by S, we can calculate that by substituting our values we have negative S prime which is negative three m divided by negative uh divided by S which is 0.3 m. And calculating this, we will get the magnification factor m of the phone screen to then be equals to negative 10. Next we want to calculate the area of the smartphone to then uh be to then be used in our next calculation to calculate the minimum area of the whiteboard based on that and the magnification of our um smartphone. So let's do that. So the area of the smartphone A will be calculated by multiplying the width and the length of the smartphone which is given in the problem statement to be five centimeter and 10 centimeter. So converting that into meter, we will then get 0.5 m and 0.10 m. So five centimeter will be converted into 0.5 m and centimeters will be converted into 0.1 m. And in doing this calculation, the area will then be equals to 5.0 times 10 to the power of negative three m squared. So earlier, we have calculated the magnification and also the area of the smartphone. So now we can calculate the minimum area of the white board. So a white board will then be equal to first, we want to uh essentially use the dimensions of the smartphone multiplied by the magnification. So that will be 0.5 m multiplied by the magnification value in absolute value which is going to be 10 and then multiply that with 0.1 m multiply what by the absolute value of the magnification as well, which is going to be 10 and calculating this, the area of the white board will then come out to a value of 0.5 m square. So then looking at the options available in the problem statement, we have to find uh the dimensions of the white board that will fit the area or that will equal to the area of 0.5 m squared. So in this case, the only uh the only option that will fit that is going to be option C So evaluating option C we will have the area of the white board to then be equal to. I'm gonna indicate that with a whiteboard prime that will be uh equal to 0.50 m multiplied by 1.0 m which will uh resulted in the area to whiteboard to be 0.5 m squared. So there we have it option C will be the answer to this particular practice problem with the area of the white board being uh giving 0.5 m squared as what we have calculated with the dimensions with the width being 0. m and the length being one m. So that will be it for this video. If you guys still have any sort of confusion, please feel free to check out our other lesson videos on similar topics available on our website. But other than that, that will be it for this one. Thank you.