So specifically we have a problem here. Now an object is placed 12 centimeters in front of a concave mirror with a focal length of 8 centimeters. So from a mirror you have your focal length half the radius of the sphere. So just beyond or a little beyond your just over your focal length is the object.
is where the object is placed. Okay, 12 centimeters away from the mirror. So from this, you're going to calculate the position of the image or you may need to say the distance of the image from the mirror. Second, the magnification of the image.
How many times the image is magnified or is it magnified or reduced in size? And the third, from your results in numbers one and two, from those numbers, you're going to interpret these numbers into the nature of the image, whether it's real image or virtual image. You will also be able to identify whether the image is inverted or upright.
And you'll also be going to identify the size of the image. Is it magnified, reduced, or same size of the object? Okay, so let's find out. the answers on these items okay so starting okay i mean we have this uh we we already made a ray diagram of this particular problem so as you notice that object placed 12 centimeters in front of a concave mirror and your focal length is only 8 centimeters so that is why the position of the object or the the purple arrow is just over or just over the focal length okay so we already know that at some point the image is bigger than the object we already know that the distance of the image is farther than 12 centimeters okay but we are not accurate about the specific distance and the specific height of the object.
So that is where we will going to use our mirror equation. Okay, so step one, using the mirror equation to find the image distance. So the mirror equation, as I presented earlier, is 1 over f is equal to 1 over d o.
plus 1 over d i f means focal length of focus d o means distance of the object plus d i meaning the distance of the image okay so we are given that the focal length is already 8 centimeters and the distance of the object is 12 centimeters and you are asked to solve for the d i or the distance of the image So as you can see here, the equation becomes 1 over di equals 1 over f minus 1 over do. So how is this equation attained? So let's, from this equation, from the original mere equation, so we just simply transfer 1 over do on the other side, so that what is left is only 1 over di on one side. So I have 1 over di on one side. And then 1 over DO is transferred on the other side.
That's why I retain that 1 over F. Then from positive 1 over DO, it becomes negative 1 over DO because it crosses or I transfer it on the other side. So making my equation now 1 over DI equals 1 over F minus 1 over DO. Okay, so from here, we will simply substitute the values. So 1 over di equals 1 over 8. That is 1 over f minus 1 over do or 1 over 12. So as you can see, we are expressing the numbers in fraction.
So no need to be afraid with fraction. You have your help or you have your best friend with you. You can always ask help from your best friend.
and I'm talking about your calculator as your best friend. So other way to do this or other way to express or give us the answer in this is by simply getting the decimal form of the fraction which is a lot more easier when you have your calculator. So 1 divided by 8 is 0.125 and minus 1 over 12, 0.08333 something. So just simply subtract the two quotient, and you end up with 1 over di. And that 1 over di will be expressed, or this 1 over di must be, must be, you have to get its reciprocal.
So you have to invert 1 over di, making it di over 1. In the same manner, you will also do the same as you subtract 1 over 8 and 1 over 12. You also have to get its reciprocal. So the final answer then will be rounded off into a whole number. So the DI is equal to 24 centimeters. Okay, as I've mentioned.
you will only get this 24 centimeters after you get the reciprocal of the value of the eye. Okay, so the eye now is equal to 24 centimeters. Okay, so that means the distance of the image from the mirror.
So using the mirror equation, we were able to arrive at the distance. So using this, we get the distance of the image. But we are not yet, we are not able to solve yet the height of the image. So for the height of the image, we'll be using this equation here for magnification. Okay.
So the DI or distance of the image is already 24 centimeters. Next, so the magnification allows us to determine the height of the image. So by magnification, you can either use DI over DO. Okay.
So we cannot use the h i and h o since we don't have the height of the object. So we can only prefer to use the distance of the image and distance of the object to determine its magnification. So by substituting the values 24 divided by 12, the answer is, or negative d i over d o, or negative 24 over 12, the answer is negative 2. So what is the magnification?
That's negative 2. So a negative 2 means, or this number negative 2 means the image is twice the size of the object. So we do not take the negative value as a value that suggests the number is lesser than 1. Now in physics, the negative sign indicates a certain convention. In this regard, I will going to give you the conventions we will be using in solving for concave mirror images and distances. So for image distance, the answer is positive 24, meaning to say the image is formed on the same side of the mirror. So the image is formed in front of the mirror.
And since the image is in front of the mirror, this type of image is called the real image. You might be asking sir what about if the image is formed behind the mirror just like plane mirror. So if the image is formed behind the mirror then that is what you call virtual image. Okay so it's opposite from a real image and magnification here is equals to negative two as I've mentioned this means that the image is twice the size of the object. At the same time the negative value there indicates that the image is inverted okay negative value of magnification means it's it's inverted positive value for magnification means the image is not inverted or it's called upright okay so for us not to be confused about this convention i give you this list so as i mentioned positive magnification means the image is upright Well, negative magnifications means image is inverted.
And for the type of image, if it's inverted, that means it's a real image. If it's virtual, that means it is behind the mirror. Even if it's in front of the mirror, that means a real image. Okay.
Another set of convention here is the DO. So distance of the object is always positive. It's because that the object is always facing in front of the mirror. So that side is always positive.
The distance of the image is, it is positive. If it's positive, then that's called a real image. If the DI is negative, that means on.
That means behind the mirror, it means a virtual image or a negative image. And a positive focus or a positive focal length means a concave mirror. And a negative focal length means a convex mirror.
So we'll get to know more about this convex mirror as we progress with our lesson. So I am just presenting to you the convex, I mean the concave mirror. Okay, so as our final interpretation of the answer based on our mirror equation, so we are able to interpret that the image distance is 24 cm, it's positive, therefore it is a real image. The image is formed or found in front of the mirror, not behind the mirror.
Next, the magnification is negative 2, meaning it is inverted since the value is negative. So the image is not upright. It is upside down.
It's inverted. And the image is bigger in size. By how much?
That's two times the original size of the object or magnified two times. And the nature of the image is that the image is real, inverted, and magnified. Okay, I hope you're able to get and able to follow the discussion about ray diagram for concave mirror and image formation. So, as a result, what we have here is, we have that this sense of the image is 24 centimeters.
And the height of the object is twice the size of the real object. Okay, or the height of the image is twice the size that of the image. Okay, so that completes our presentation, the use of ray diagram to determine the image formation and the location, and the use of mirror equation to be able to address the accurate distances and height of the image.
Now, it's your turn. Given a concave mirror that has a focal length of 10 centimeters, object is 18 centimeters away from the mirror and the object's height is 9 centimeters or the object is 9 centimeters tall so first draw the image using ray diagram the way we did earlier the way we the way to stand earlier so follow that rule so there are only two ray of light that you're going to follow one that is parallel to the principal axis. When it hits the mirror, it will pass through the focus.
The other one is the ray of light that passing through the focus. When it hits the mirror, then it will reflect parallel to the principal axis. The intersection between these two rays of light is the location of your image. Next, you calculate the image distance and the image height you see. the meter equation.
Okay, so I hope you'll be able to answer this and write your answer in a one-half crosswise piece of paper and that will be submitted during our science session. Okay, so that will be all for now. Always remember, keep going, keep growing.
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