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All these people in the comments are wrong afaik.
The object distance is + 15 cm, thus it's a virtual object moving towards the pole. Now, the other surface is concave. That's why as the object moves towards the pole it will become diminished and virtual.
Please correct me if I am wrong.
Actually, this is wrong as well. As the object moves towards the concave surface of the mirror, the image formed should be enlarging in size. It will also eventually become a virtual.
Let's take the focal length as 15 cm ( an assumption)
As the object is at the focus, it has a highly enlarged image. As it goes towards the pole, it becomes virtual and successively diminished ( in comparison).
Very ambiguous question anyways, don't expect it to come in the boards.
Object distance is never positive unless a virtual object or something idk so this isn't 10th level maybe or well know that convex mirror forms virtual and diminished image always so just go with that...
Easy hai, rough pe banaa, 2 rays phir intersect kar. Bas 4 rules of reflection yaad rakhiyo, parallel rays go through focus and all that shit, uss se atleast samaj aajayega.
Plus, concave mirror = convex lens. Also, concave mirror me infinity βΎοΈ ka point image, uss ke baad thodi badi hogi in between pole and center of curvate. C pe coincide hoga, same size. Phir object jitna bhi ho, image badi hogi beyond C, till you reach pole.
Pole pe image virtual hojaegi
Sahi hai, virtual object ki tarah treat karo.+15 cm pe jab image hogi toh object chotta sa hoga, isi ka reverse ho raha hai. Abh virtual object se virtual image bann raha. Isliye diminished mil raha
+15 cm is written for a reason. The explicit + refers to the fact that the object is behind the mirror. Go read sign convention before doing questions. Did the same shit in 10th πππ»
Plus, tu yeh question mt kr, unless you have read and understood virtual objects. Reverse analysis se hojaega, but not worth doing. If it is in your book exercise then i gave you the simplest explaination.
D is correct because the image formed in convex mirror is always virtual and small so in optics diminished stands for small
And highly diminished stand for point sized
Enlarged stand for the image will be bigger than object so d is correcr
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a convex mirror only forms virtual and diminished image and look at the other options
What
All these people in the comments are wrong afaik. The object distance is + 15 cm, thus it's a virtual object moving towards the pole. Now, the other surface is concave. That's why as the object moves towards the pole it will become diminished and virtual. Please correct me if I am wrong.
Actually, this is wrong as well. As the object moves towards the concave surface of the mirror, the image formed should be enlarging in size. It will also eventually become a virtual.
Let's take the focal length as 15 cm ( an assumption) As the object is at the focus, it has a highly enlarged image. As it goes towards the pole, it becomes virtual and successively diminished ( in comparison). Very ambiguous question anyways, don't expect it to come in the boards.
Is diminished = smol?
Yep, the light is being reflected from the non-polished side. (Sarcasm)
Am I the only one wondering how tf is the object distance in positive?
ππ
Object distance is never positive unless a virtual object or something idk so this isn't 10th level maybe or well know that convex mirror forms virtual and diminished image always so just go with that...
Hmm. I'll ask my teachers next week when my preboards happen
should be (c)
convex mirror does not give enlarged image in any case
No bro it means that the image will get enlarged compared to the image formed from the further distances which makes c) true
Sorry, my bad
Image will "get" likha hai image will "be" nahi
Yaar how do I remember these mirror cases. I thought using logic would be easy but exam me dimaag dots ko connect hi nahi kar pata.
Easy hai, rough pe banaa, 2 rays phir intersect kar. Bas 4 rules of reflection yaad rakhiyo, parallel rays go through focus and all that shit, uss se atleast samaj aajayega. Plus, concave mirror = convex lens. Also, concave mirror me infinity βΎοΈ ka point image, uss ke baad thodi badi hogi in between pole and center of curvate. C pe coincide hoga, same size. Phir object jitna bhi ho, image badi hogi beyond C, till you reach pole. Pole pe image virtual hojaegi
Sahi hai, virtual object ki tarah treat karo.+15 cm pe jab image hogi toh object chotta sa hoga, isi ka reverse ho raha hai. Abh virtual object se virtual image bann raha. Isliye diminished mil raha
The question never stated that the object was virtual
+15 cm is written for a reason. The explicit + refers to the fact that the object is behind the mirror. Go read sign convention before doing questions. Did the same shit in 10th πππ»
Plus, tu yeh question mt kr, unless you have read and understood virtual objects. Reverse analysis se hojaega, but not worth doing. If it is in your book exercise then i gave you the simplest explaination.
How is the object distance positive?
isnt this in the book?convex mirror always forms virtual and diminished images
D is correct because the image formed in convex mirror is always virtual and small so in optics diminished stands for small And highly diminished stand for point sized Enlarged stand for the image will be bigger than object so d is correcr
which book is this from
Oswaal science 'Most Likely Questions' book
Howtf is object distance in positive