From 48daf84d28689f3ca8eb97d973b65198edfc90df Mon Sep 17 00:00:00 2001 From: hussein Date: Thu, 25 Aug 2022 15:45:09 +0200 Subject: [PATCH 01/10] try to fix the eq Angular res --- .../links/.ipynb_checkpoints/Angular resolution-checkpoint.md | 2 +- resources/links/Angular resolution.md | 2 +- 2 files changed, 2 insertions(+), 2 deletions(-) diff --git a/resources/links/.ipynb_checkpoints/Angular resolution-checkpoint.md b/resources/links/.ipynb_checkpoints/Angular resolution-checkpoint.md index 616a3fb..67db42a 100644 --- a/resources/links/.ipynb_checkpoints/Angular resolution-checkpoint.md +++ b/resources/links/.ipynb_checkpoints/Angular resolution-checkpoint.md @@ -13,7 +13,7 @@ It seems that in astronomy, there is a relationship between the image resolution $\theta$ and the wavelength $\lambda$ giving by - $$ \theta = 1.22 \frac{\lambda} {D} $$ + $$\theta = 1.22 \frac{\lambda} {D}$$ - Where $D$ is the diameter of the telescope. - to get better resolution (small $\theta$), we need: diff --git a/resources/links/Angular resolution.md b/resources/links/Angular resolution.md index 616a3fb..329fefb 100644 --- a/resources/links/Angular resolution.md +++ b/resources/links/Angular resolution.md @@ -13,7 +13,7 @@ It seems that in astronomy, there is a relationship between the image resolution $\theta$ and the wavelength $\lambda$ giving by - $$ \theta = 1.22 \frac{\lambda} {D} $$ + $$\theta = 1.22 \frac{\lambda} {D}$$ - Where $D$ is the diameter of the telescope. - to get better resolution (small $\theta$), we need: From 0cdb8734851a455b3069a79028d4320de1e4600d Mon Sep 17 00:00:00 2001 From: hussein Date: Thu, 25 Aug 2022 15:47:14 +0200 Subject: [PATCH 02/10] s --- resources/links/Angular resolution.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/resources/links/Angular resolution.md b/resources/links/Angular resolution.md index 329fefb..e220636 100644 --- a/resources/links/Angular resolution.md +++ b/resources/links/Angular resolution.md @@ -13,7 +13,7 @@ It seems that in astronomy, there is a relationship between the image resolution $\theta$ and the wavelength $\lambda$ giving by - $$\theta = 1.22 \frac{\lambda} {D}$$ +$$\theta = 1.22 \frac{\lambda} {D}$$ - Where $D$ is the diameter of the telescope. - to get better resolution (small $\theta$), we need: From c908d5b1cfa71950ebccc1f2af148831fed381dc Mon Sep 17 00:00:00 2001 From: hussein Date: Thu, 25 Aug 2022 15:48:17 +0200 Subject: [PATCH 03/10] still fix it --- .../links/.ipynb_checkpoints/Angular resolution-checkpoint.md | 2 +- resources/links/Angular resolution.md | 2 +- 2 files changed, 2 insertions(+), 2 deletions(-) diff --git a/resources/links/.ipynb_checkpoints/Angular resolution-checkpoint.md b/resources/links/.ipynb_checkpoints/Angular resolution-checkpoint.md index 67db42a..d83233c 100644 --- a/resources/links/.ipynb_checkpoints/Angular resolution-checkpoint.md +++ b/resources/links/.ipynb_checkpoints/Angular resolution-checkpoint.md @@ -13,7 +13,7 @@ It seems that in astronomy, there is a relationship between the image resolution $\theta$ and the wavelength $\lambda$ giving by - $$\theta = 1.22 \frac{\lambda} {D}$$ +$\theta = 1.22 \frac{\lambda} {D}$ - Where $D$ is the diameter of the telescope. - to get better resolution (small $\theta$), we need: diff --git a/resources/links/Angular resolution.md b/resources/links/Angular resolution.md index e220636..d83233c 100644 --- a/resources/links/Angular resolution.md +++ b/resources/links/Angular resolution.md @@ -13,7 +13,7 @@ It seems that in astronomy, there is a relationship between the image resolution $\theta$ and the wavelength $\lambda$ giving by -$$\theta = 1.22 \frac{\lambda} {D}$$ +$\theta = 1.22 \frac{\lambda} {D}$ - Where $D$ is the diameter of the telescope. - to get better resolution (small $\theta$), we need: From bcd32a75313281eff9cf70305dd893c725a187bd Mon Sep 17 00:00:00 2001 From: Eslam Hussein <31181109+eahussein@users.noreply.github.com> Date: Thu, 25 Aug 2022 15:50:41 +0200 Subject: [PATCH 04/10] Update Angular resolution.md --- resources/links/Angular resolution.md | 1 + 1 file changed, 1 insertion(+) diff --git a/resources/links/Angular resolution.md b/resources/links/Angular resolution.md index d83233c..c8b994f 100644 --- a/resources/links/Angular resolution.md +++ b/resources/links/Angular resolution.md @@ -14,6 +14,7 @@ It seems that in astronomy, there is a relationship between the image resolution $\theta$ and the wavelength $\lambda$ giving by $\theta = 1.22 \frac{\lambda} {D}$ +$$\int_\Omega \nabla u \cdot \nabla v~dx = \int_\Omega fv~dx$$ - Where $D$ is the diameter of the telescope. - to get better resolution (small $\theta$), we need: From b2f4a058b17c52e732a4dd97f263c79d405a1320 Mon Sep 17 00:00:00 2001 From: Eslam Hussein <31181109+eahussein@users.noreply.github.com> Date: Thu, 25 Aug 2022 15:51:08 +0200 Subject: [PATCH 05/10] Update Angular resolution.md --- resources/links/Angular resolution.md | 5 +++-- 1 file changed, 3 insertions(+), 2 deletions(-) diff --git a/resources/links/Angular resolution.md b/resources/links/Angular resolution.md index c8b994f..c9f5d9f 100644 --- a/resources/links/Angular resolution.md +++ b/resources/links/Angular resolution.md @@ -13,8 +13,9 @@ It seems that in astronomy, there is a relationship between the image resolution $\theta$ and the wavelength $\lambda$ giving by -$\theta = 1.22 \frac{\lambda} {D}$ -$$\int_\Omega \nabla u \cdot \nabla v~dx = \int_\Omega fv~dx$$ + $$\theta = 1.22 \frac{\lambda} {D}$$ + + $$\int_\Omega \nabla u \cdot \nabla v~dx = \int_\Omega fv~dx$$ - Where $D$ is the diameter of the telescope. - to get better resolution (small $\theta$), we need: From 66e1601de8052f755804fd79a5e46c8cbccc5146 Mon Sep 17 00:00:00 2001 From: Eslam Hussein <31181109+eahussein@users.noreply.github.com> Date: Thu, 25 Aug 2022 15:53:54 +0200 Subject: [PATCH 06/10] Update Angular resolution.md --- resources/links/Angular resolution.md | 5 +++-- 1 file changed, 3 insertions(+), 2 deletions(-) diff --git a/resources/links/Angular resolution.md b/resources/links/Angular resolution.md index c9f5d9f..ab92e46 100644 --- a/resources/links/Angular resolution.md +++ b/resources/links/Angular resolution.md @@ -13,9 +13,10 @@ It seems that in astronomy, there is a relationship between the image resolution $\theta$ and the wavelength $\lambda$ giving by - $$\theta = 1.22 \frac{\lambda} {D}$$ +**The Cauchy-Schwarz Inequality** + +$$\left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)$$ - $$\int_\Omega \nabla u \cdot \nabla v~dx = \int_\Omega fv~dx$$ - Where $D$ is the diameter of the telescope. - to get better resolution (small $\theta$), we need: From 68e3252758b5637d6be5e409f9d429f46a5fe2ee Mon Sep 17 00:00:00 2001 From: Eslam Hussein <31181109+eahussein@users.noreply.github.com> Date: Thu, 25 Aug 2022 15:55:56 +0200 Subject: [PATCH 07/10] Update Angular resolution.md --- resources/links/Angular resolution.md | 3 +-- 1 file changed, 1 insertion(+), 2 deletions(-) diff --git a/resources/links/Angular resolution.md b/resources/links/Angular resolution.md index ab92e46..9400bee 100644 --- a/resources/links/Angular resolution.md +++ b/resources/links/Angular resolution.md @@ -13,9 +13,8 @@ It seems that in astronomy, there is a relationship between the image resolution $\theta$ and the wavelength $\lambda$ giving by -**The Cauchy-Schwarz Inequality** -$$\left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)$$ +$$\sqrt{3x-1}+(1+x)^2$$ - Where $D$ is the diameter of the telescope. From 53b78f849c94be9223198844c43f6e93876b636f Mon Sep 17 00:00:00 2001 From: Eslam Hussein <31181109+eahussein@users.noreply.github.com> Date: Thu, 25 Aug 2022 15:56:10 +0200 Subject: [PATCH 08/10] Update Angular resolution.md --- resources/links/Angular resolution.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/resources/links/Angular resolution.md b/resources/links/Angular resolution.md index 9400bee..0b8c964 100644 --- a/resources/links/Angular resolution.md +++ b/resources/links/Angular resolution.md @@ -14,7 +14,7 @@ It seems that in astronomy, there is a relationship between the image resolution $\theta$ and the wavelength $\lambda$ giving by -$$\sqrt{3x-1}+(1+x)^2$$ +$$\sqrt{3x-1} = (1+x)^2$$ - Where $D$ is the diameter of the telescope. From 6c2cc9754d08cae4d453a5299e5bc1035731fa4b Mon Sep 17 00:00:00 2001 From: Eslam Hussein <31181109+eahussein@users.noreply.github.com> Date: Thu, 25 Aug 2022 15:56:41 +0200 Subject: [PATCH 09/10] Update Angular resolution.md --- resources/links/Angular resolution.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/resources/links/Angular resolution.md b/resources/links/Angular resolution.md index 0b8c964..114e0b3 100644 --- a/resources/links/Angular resolution.md +++ b/resources/links/Angular resolution.md @@ -14,7 +14,7 @@ It seems that in astronomy, there is a relationship between the image resolution $\theta$ and the wavelength $\lambda$ giving by -$$\sqrt{3x-1} = (1+x)^2$$ +$$\theta = 1.22 \frac{\lambda} {D}$$ - Where $D$ is the diameter of the telescope. From 0b8aca8b89a71aaeb02a4f5cbfb029322f9a89a1 Mon Sep 17 00:00:00 2001 From: Eslam Hussein <31181109+eahussein@users.noreply.github.com> Date: Thu, 25 Aug 2022 15:57:35 +0200 Subject: [PATCH 10/10] Update Angular resolution.md --- resources/links/Angular resolution.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/resources/links/Angular resolution.md b/resources/links/Angular resolution.md index 114e0b3..cb55d81 100644 --- a/resources/links/Angular resolution.md +++ b/resources/links/Angular resolution.md @@ -11,7 +11,7 @@ **_SO THE GOAL BECOMES HOW TO MAKE IMAGES WITH SMALL ANGULAR RESOLUTIONS._** -It seems that in astronomy, there is a relationship between the image resolution $\theta$ and the wavelength $\lambda$ giving by +It seems that in astronomy, there is a relationship between the image resolution $\theta$ and the wavelength $\lambda$ giving by: $$\theta = 1.22 \frac{\lambda} {D}$$