Practical: removing parallax between image and pin, which way do I move the pin if it moves with my eye?
Neel Chatterjee12th Sample
21 Aug 2026 · 1 min read
Practical exam coming up and I want to actually understand it, not just copy the observation table. We find the focal length of a concave mirror (and a convex lens) by the u–v method. I place the object pin, get a real inverted image, and place an image pin where the image appears to be. To check, I move my head left and right and see whether the tip of the image and the tip of the pin move together.
My trouble: sometimes the pin seems to move with my head and sometimes against it. My lab teacher said "move the pin accordingly" and walked off.
- If the pin moves in the same direction as my eye, is the pin too near or too far from the image?
- Why do we keep the eye at least 25–30 cm away from the image?
- Instead of calculating f from each reading, the lab manual plots a graph. What exactly do I plot and how do I read f?
- What is the "index correction" and do I need it?
Chess taught me to ask for the rule, not the vibes.
7 answers & replies
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[+][–]Rohan Sharma11thSample· 1mo ago
if pin moves with ur eye then pin is too near, push it away from you. if it moves opposite then bring it closer. thats what i remember from the demo
[+][–]Aman YadavDropperSample· 1mo ago
Rohan, it is the other way round. Rule: of two objects in the same line of sight, the farther one appears to move with your eye relative to the nearer one. The nearer one seems to move against.
So: pin moves with your eye (relative to the image) → pin is farther from you than the image → bring it closer. Pin moves against → pin is nearer than the image → push it away. Repeat till the two tips stay together when you move your head. That is "no parallax": tip to tip.
[+][–]Neel Chatterjee12thSample· 1mo agoAuthor
Got it, "farther one goes with the eye" is the rule. And I wasn't aware of the 25 cm reason, that's actually cool. Will re-do my table with the 1/u–1/v graph.
[+][–]Rohan Sharma11thSample· 1mo ago
bruh i had it backwards, glad this came up before the exam 😭
[+][–]Tanvi Deshmukh12thSample· 1mo ago
Why 25 cm!! The image in this experiment is a real image, formed in front of the mirror in the air, and your eye has to look at it like an actual object. The eye cannot focus clearly on anything nearer than about 25 cm (the least distance of distinct vision). So if your eye is closer, the image looks blurry and you'll place the pin wrongly. Keep your eye 30 cm or more from the image and look slightly to the side so the real image is seen properly.
[+][–]Vivek Mishra12thSample· 1mo ago
For the graph. Take readings for different object positions and plot (y-axis) against (x-axis). From you get a straight line with a slope of −1. The line cuts both axes. The intercept on each axis (taken with sign) is . So = 1 / intercept; take the average of the two intercepts. For a concave mirror u and v are negative, so the intercepts are negative, and that gives a negative f.
Graph averages the errors out instead of trusting one bad reading, that's why they ask for it.
[+][–]Farhan ShaikhDropperSample· 1mo ago
Index correction: you are measuring the distance from the pole of the mirror to the tip of the pin, but the scale and the pin holder aren't always aligned. So the measured distance is a bit off. To find the error you place the pin tip touching the mirror's pole and read the scale; the difference from the true distance is the index correction, and you add/subtract it to every u and v. It's usually a small value, but the examiner likes to see it written in the observations.
